Related
I have two DataFrames which I want to merge based on a column. However, due to alternate spellings, different number of spaces, absence/presence of diacritical marks, I would like to be able to merge as long as they are similar to one another.
Any similarity algorithm will do (soundex, Levenshtein, difflib's).
Say one DataFrame has the following data:
df1 = DataFrame([[1],[2],[3],[4],[5]], index=['one','two','three','four','five'], columns=['number'])
number
one 1
two 2
three 3
four 4
five 5
df2 = DataFrame([['a'],['b'],['c'],['d'],['e']], index=['one','too','three','fours','five'], columns=['letter'])
letter
one a
too b
three c
fours d
five e
Then I want to get the resulting DataFrame
number letter
one 1 a
two 2 b
three 3 c
four 4 d
five 5 e
Similar to #locojay suggestion, you can apply difflib's get_close_matches to df2's index and then apply a join:
In [23]: import difflib
In [24]: difflib.get_close_matches
Out[24]: <function difflib.get_close_matches>
In [25]: df2.index = df2.index.map(lambda x: difflib.get_close_matches(x, df1.index)[0])
In [26]: df2
Out[26]:
letter
one a
two b
three c
four d
five e
In [31]: df1.join(df2)
Out[31]:
number letter
one 1 a
two 2 b
three 3 c
four 4 d
five 5 e
.
If these were columns, in the same vein you could apply to the column then merge:
df1 = DataFrame([[1,'one'],[2,'two'],[3,'three'],[4,'four'],[5,'five']], columns=['number', 'name'])
df2 = DataFrame([['a','one'],['b','too'],['c','three'],['d','fours'],['e','five']], columns=['letter', 'name'])
df2['name'] = df2['name'].apply(lambda x: difflib.get_close_matches(x, df1['name'])[0])
df1.merge(df2)
Using fuzzywuzzy
Since there are no examples with the fuzzywuzzy package, here's a function I wrote which will return all matches based on a threshold you can set as a user:
Example datframe
df1 = pd.DataFrame({'Key':['Apple', 'Banana', 'Orange', 'Strawberry']})
df2 = pd.DataFrame({'Key':['Aple', 'Mango', 'Orag', 'Straw', 'Bannanna', 'Berry']})
# df1
Key
0 Apple
1 Banana
2 Orange
3 Strawberry
# df2
Key
0 Aple
1 Mango
2 Orag
3 Straw
4 Bannanna
5 Berry
Function for fuzzy matching
def fuzzy_merge(df_1, df_2, key1, key2, threshold=90, limit=2):
"""
:param df_1: the left table to join
:param df_2: the right table to join
:param key1: key column of the left table
:param key2: key column of the right table
:param threshold: how close the matches should be to return a match, based on Levenshtein distance
:param limit: the amount of matches that will get returned, these are sorted high to low
:return: dataframe with boths keys and matches
"""
s = df_2[key2].tolist()
m = df_1[key1].apply(lambda x: process.extract(x, s, limit=limit))
df_1['matches'] = m
m2 = df_1['matches'].apply(lambda x: ', '.join([i[0] for i in x if i[1] >= threshold]))
df_1['matches'] = m2
return df_1
Using our function on the dataframes: #1
from fuzzywuzzy import fuzz
from fuzzywuzzy import process
fuzzy_merge(df1, df2, 'Key', 'Key', threshold=80)
Key matches
0 Apple Aple
1 Banana Bannanna
2 Orange Orag
3 Strawberry Straw, Berry
Using our function on the dataframes: #2
df1 = pd.DataFrame({'Col1':['Microsoft', 'Google', 'Amazon', 'IBM']})
df2 = pd.DataFrame({'Col2':['Mcrsoft', 'gogle', 'Amason', 'BIM']})
fuzzy_merge(df1, df2, 'Col1', 'Col2', 80)
Col1 matches
0 Microsoft Mcrsoft
1 Google gogle
2 Amazon Amason
3 IBM
Installation:
Pip
pip install fuzzywuzzy
Anaconda
conda install -c conda-forge fuzzywuzzy
I have written a Python package which aims to solve this problem:
pip install fuzzymatcher
You can find the repo here and docs here.
Basic usage:
Given two dataframes df_left and df_right, which you want to fuzzy join, you can write the following:
from fuzzymatcher import link_table, fuzzy_left_join
# Columns to match on from df_left
left_on = ["fname", "mname", "lname", "dob"]
# Columns to match on from df_right
right_on = ["name", "middlename", "surname", "date"]
# The link table potentially contains several matches for each record
fuzzymatcher.link_table(df_left, df_right, left_on, right_on)
Or if you just want to link on the closest match:
fuzzymatcher.fuzzy_left_join(df_left, df_right, left_on, right_on)
I would use Jaro-Winkler, because it is one of the most performant and accurate approximate string matching algorithms currently available [Cohen, et al.], [Winkler].
This is how I would do it with Jaro-Winkler from the jellyfish package:
def get_closest_match(x, list_strings):
best_match = None
highest_jw = 0
for current_string in list_strings:
current_score = jellyfish.jaro_winkler(x, current_string)
if(current_score > highest_jw):
highest_jw = current_score
best_match = current_string
return best_match
df1 = pandas.DataFrame([[1],[2],[3],[4],[5]], index=['one','two','three','four','five'], columns=['number'])
df2 = pandas.DataFrame([['a'],['b'],['c'],['d'],['e']], index=['one','too','three','fours','five'], columns=['letter'])
df2.index = df2.index.map(lambda x: get_closest_match(x, df1.index))
df1.join(df2)
Output:
number letter
one 1 a
two 2 b
three 3 c
four 4 d
five 5 e
For a general approach: fuzzy_merge
For a more general scenario in which we want to merge columns from two dataframes which contain slightly different strings, the following function uses difflib.get_close_matches along with merge in order to mimic the functionality of pandas' merge but with fuzzy matching:
import difflib
def fuzzy_merge(df1, df2, left_on, right_on, how='inner', cutoff=0.6):
df_other= df2.copy()
df_other[left_on] = [get_closest_match(x, df1[left_on], cutoff)
for x in df_other[right_on]]
return df1.merge(df_other, on=left_on, how=how)
def get_closest_match(x, other, cutoff):
matches = difflib.get_close_matches(x, other, cutoff=cutoff)
return matches[0] if matches else None
Here are some use cases with two sample dataframes:
print(df1)
key number
0 one 1
1 two 2
2 three 3
3 four 4
4 five 5
print(df2)
key_close letter
0 three c
1 one a
2 too b
3 fours d
4 a very different string e
With the above example, we'd get:
fuzzy_merge(df1, df2, left_on='key', right_on='key_close')
key number key_close letter
0 one 1 one a
1 two 2 too b
2 three 3 three c
3 four 4 fours d
And we could do a left join with:
fuzzy_merge(df1, df2, left_on='key', right_on='key_close', how='left')
key number key_close letter
0 one 1 one a
1 two 2 too b
2 three 3 three c
3 four 4 fours d
4 five 5 NaN NaN
For a right join, we'd have all non-matching keys in the left dataframe to None:
fuzzy_merge(df1, df2, left_on='key', right_on='key_close', how='right')
key number key_close letter
0 one 1.0 one a
1 two 2.0 too b
2 three 3.0 three c
3 four 4.0 fours d
4 None NaN a very different string e
Also note that difflib.get_close_matches will return an empty list if no item is matched within the cutoff. In the shared example, if we change the last index in df2 to say:
print(df2)
letter
one a
too b
three c
fours d
a very different string e
We'd get an index out of range error:
df2.index.map(lambda x: difflib.get_close_matches(x, df1.index)[0])
IndexError: list index out of range
In order to solve this the above function get_closest_match will return the closest match by indexing the list returned by difflib.get_close_matches only if it actually contains any matches.
http://pandas.pydata.org/pandas-docs/dev/merging.html does not have a hook function to do this on the fly. Would be nice though...
I would just do a separate step and use difflib getclosest_matches to create a new column in one of the 2 dataframes and the merge/join on the fuzzy matched column
I used Fuzzymatcher package and this worked well for me. Visit this link for more details on this.
use the below command to install
pip install fuzzymatcher
Below is the sample Code (already submitted by RobinL above)
from fuzzymatcher import link_table, fuzzy_left_join
# Columns to match on from df_left
left_on = ["fname", "mname", "lname", "dob"]
# Columns to match on from df_right
right_on = ["name", "middlename", "surname", "date"]
# The link table potentially contains several matches for each record
fuzzymatcher.link_table(df_left, df_right, left_on, right_on)
Errors you may get
ZeroDivisionError: float division by zero---> Refer to this
link to resolve it
OperationalError: No Such Module:fts4 --> downlaod the sqlite3.dll
from here and replace the DLL file in your python or anaconda
DLLs folder.
Pros :
Works faster. In my case, I compared one dataframe with 3000 rows with anohter dataframe with 170,000 records . This also uses SQLite3 search across text. So faster than many
Can check across multiple columns and 2 dataframes. In my case, I was looking for closest match based on address and company name. Sometimes, company name might be same but address is the good thing to check too.
Gives you score for all the closest matches for the same record. you choose whats the cutoff score.
cons:
Original package installation is buggy
Required C++ and visual studios installed too
Wont work for 64 bit anaconda/Python
There is a package called fuzzy_pandas that can use levenshtein, jaro, metaphone and bilenco methods. With some great examples here
import pandas as pd
import fuzzy_pandas as fpd
df1 = pd.DataFrame({'Key':['Apple', 'Banana', 'Orange', 'Strawberry']})
df2 = pd.DataFrame({'Key':['Aple', 'Mango', 'Orag', 'Straw', 'Bannanna', 'Berry']})
results = fpd.fuzzy_merge(df1, df2,
left_on='Key',
right_on='Key',
method='levenshtein',
threshold=0.6)
results.head()
Key Key
0 Apple Aple
1 Banana Bannanna
2 Orange Orag
As a heads up, this basically works, except if no match is found, or if you have NaNs in either column. Instead of directly applying get_close_matches, I found it easier to apply the following function. The choice of NaN replacements will depend a lot on your dataset.
def fuzzy_match(a, b):
left = '1' if pd.isnull(a) else a
right = b.fillna('2')
out = difflib.get_close_matches(left, right)
return out[0] if out else np.NaN
You can use d6tjoin for that
import d6tjoin.top1
d6tjoin.top1.MergeTop1(df1.reset_index(),df2.reset_index(),
fuzzy_left_on=['index'],fuzzy_right_on=['index']).merge()['merged']
index number index_right letter
0 one 1 one a
1 two 2 too b
2 three 3 three c
3 four 4 fours d
4 five 5 five e
It has a variety of additional features such as:
check join quality, pre and post join
customize similarity function, eg edit distance vs hamming distance
specify max distance
multi-core compute
For details see
MergeTop1 examples - Best match join examples notebook
PreJoin examples - Examples for diagnosing join problems
I have used fuzzywuzz in a very minimal way whilst matching the existing behaviour and keywords of merge in pandas.
Just specify your accepted threshold for matching (between 0 and 100):
from fuzzywuzzy import process
def fuzzy_merge(df, df2, on=None, left_on=None, right_on=None, how='inner', threshold=80):
def fuzzy_apply(x, df, column, threshold=threshold):
if type(x)!=str:
return None
match, score, *_ = process.extract(x, df[column], limit=1)[0]
if score >= threshold:
return match
else:
return None
if on is not None:
left_on = on
right_on = on
# create temp column as the best fuzzy match (or None!)
df2['tmp'] = df2[right_on].apply(
fuzzy_apply,
df=df,
column=left_on,
threshold=threshold
)
merged_df = df.merge(df2, how=how, left_on=left_on, right_on='tmp')
del merged_df['tmp']
return merged_df
Try it out using the example data:
df1 = pd.DataFrame({'Key':['Apple', 'Banana', 'Orange', 'Strawberry']})
df2 = pd.DataFrame({'Key':['Aple', 'Mango', 'Orag', 'Straw', 'Bannanna', 'Berry']})
fuzzy_merge(df, df2, on='Key', threshold=80)
Using thefuzz
Using SeatGeek's great package thefuzz, which makes use of Levenshtein distance. This works with data held in columns. It adds matches as rows rather than columns, to preserve a tidy dataset, and allows additional columns to be easily pulled through to the output dataframe.
Sample data
df1 = pd.DataFrame({'col_a':['one','two','three','four','five'], 'col_b':[1, 2, 3, 4, 5]})
col_a col_b
0 one 1
1 two 2
2 three 3
3 four 4
4 five 5
df2 = pd.DataFrame({'col_a':['one','too','three','fours','five'], 'col_b':['a','b','c','d','e']})
col_a col_b
0 one a
1 too b
2 three c
3 fours d
4 five e
Function used to do the matching
def fuzzy_match(
df_left, df_right, column_left, column_right, threshold=90, limit=1
):
# Create a series
series_matches = df_left[column_left].apply(
lambda x: process.extract(x, df_right[column_right], limit=limit) # Creates a series with id from df_left and column name _column_left_, with _limit_ matches per item
)
# Convert matches to a tidy dataframe
df_matches = series_matches.to_frame()
df_matches = df_matches.explode(column_left) # Convert list of matches to rows
df_matches[
['match_string', 'match_score', 'df_right_id']
] = pd.DataFrame(df_matches[column_left].tolist(), index=df_matches.index) # Convert match tuple to columns
df_matches.drop(column_left, axis=1, inplace=True) # Drop column of match tuples
# Reset index, as in creating a tidy dataframe we've introduced multiple rows per id, so that no longer functions well as the index
if df_matches.index.name:
index_name = df_matches.index.name # Stash index name
else:
index_name = 'index' # Default used by pandas
df_matches.reset_index(inplace=True)
df_matches.rename(columns={index_name: 'df_left_id'}, inplace=True) # The previous index has now become a column: rename for ease of reference
# Drop matches below threshold
df_matches.drop(
df_matches.loc[df_matches['match_score'] < threshold].index,
inplace=True
)
return df_matches
Use function and merge data
import pandas as pd
from thefuzz import process
df_matches = fuzzy_match(
df1,
df2,
'col_a',
'col_a',
threshold=60,
limit=1
)
df_output = df1.merge(
df_matches,
how='left',
left_index=True,
right_on='df_left_id'
).merge(
df2,
how='left',
left_on='df_right_id',
right_index=True,
suffixes=['_df1', '_df2']
)
df_output.set_index('df_left_id', inplace=True) # For some reason the first merge operation wrecks the dataframe's index. Recreated from the value we have in the matches lookup table
df_output = df_output[['col_a_df1', 'col_b_df1', 'col_b_df2']] # Drop columns used in the matching
df_output.index.name = 'id'
id col_a_df1 col_b_df1 col_b_df2
0 one 1 a
1 two 2 b
2 three 3 c
3 four 4 d
4 five 5 e
Tip: Fuzzy matching using thefuzz is much quicker if you optionally install the python-Levenshtein package too.
For more complex use cases to match rows with many columns you can use recordlinkage package. recordlinkage provides all the tools to fuzzy match rows between pandas data frames which helps to deduplicate your data when merging. I have written a detailed article about the package here
if the join axis is numeric this could also be used to match indexes with a specified tolerance:
def fuzzy_left_join(df1, df2, tol=None):
index1 = df1.index.values
index2 = df2.index.values
diff = np.abs(index1.reshape((-1, 1)) - index2)
mask_j = np.argmin(diff, axis=1) # min. of each column
mask_i = np.arange(mask_j.shape[0])
df1_ = df1.iloc[mask_i]
df2_ = df2.iloc[mask_j]
if tol is not None:
mask = np.abs(df2_.index.values - df1_.index.values) <= tol
df1_ = df1_.loc[mask]
df2_ = df2_.loc[mask]
df2_.index = df1_.index
out = pd.concat([df1_, df2_], axis=1)
return out
TheFuzz is the new version of a fuzzywuzzy
In order to fuzzy-join string-elements in two big tables you can do this:
Use apply to go row by row
Use swifter to parallel, speed up and visualize default apply function (with colored progress bar)
Use OrderedDict from collections to get rid of duplicates in the output of merge and keep the initial order
Increase limit in thefuzz.process.extract to see more options for merge (stored in a list of tuples with % of similarity)
'*' You can use thefuzz.process.extractOne instead of thefuzz.process.extract to return just one best-matched item (without specifying any limit). However, be aware that several results could have same % of similarity and you will get only one of them.
'**' Somehow the swifter takes a minute or two before starting the actual apply. If you need to process small tables you can skip this step and just use progress_apply instead
from thefuzz import process
from collections import OrderedDict
import swifter
def match(x):
matches = process.extract(x, df1, limit=6)
matches = list(OrderedDict((x, True) for x in matches).keys())
print(f'{x:20} : {matches}')
return str(matches)
df1 = df['name'].values
df2['matches'] = df2['name'].swifter.apply(lambda x: match(x))
I have a data frame having dictionary like structure. I want to only sum the values and store into new column.
Column 1 Desired Output
[{'Apple':3}, 9
{'Mango:2},
{'Peach:4}]
[{'Blue':2}, 3
{'Black':1}]
df['Desired Output'] = [sum(x) for x in df['Column 1']]
df
Assuming your Column 1 column does indeed have dictionaries (and not strings that look like dictionaries), this should do the trick:
df['Desired Output'] = df["Column 1"].apply(lambda lst: sum(sum(d.values()) for d in lst))
I load some machine learning data from a CSV file. The first 2 columns are observations and the remaining columns are features.
Currently, I do the following:
data = pandas.read_csv('mydata.csv')
which gives something like:
data = pandas.DataFrame(np.random.rand(10,5), columns = list('abcde'))
I'd like to slice this dataframe in two dataframes: one containing the columns a and b and one containing the columns c, d and e.
It is not possible to write something like
observations = data[:'c']
features = data['c':]
I'm not sure what the best method is. Do I need a pd.Panel?
By the way, I find dataframe indexing pretty inconsistent: data['a'] is permitted, but data[0] is not. On the other side, data['a':] is not permitted but data[0:] is.
Is there a practical reason for this? This is really confusing if columns are indexed by Int, given that data[0] != data[0:1]
2017 Answer - pandas 0.20: .ix is deprecated. Use .loc
See the deprecation in the docs
.loc uses label based indexing to select both rows and columns. The labels being the values of the index or the columns. Slicing with .loc includes the last element.
Let's assume we have a DataFrame with the following columns:
foo, bar, quz, ant, cat, sat, dat.
# selects all rows and all columns beginning at 'foo' up to and including 'sat'
df.loc[:, 'foo':'sat']
# foo bar quz ant cat sat
.loc accepts the same slice notation that Python lists do for both row and columns. Slice notation being start:stop:step
# slice from 'foo' to 'cat' by every 2nd column
df.loc[:, 'foo':'cat':2]
# foo quz cat
# slice from the beginning to 'bar'
df.loc[:, :'bar']
# foo bar
# slice from 'quz' to the end by 3
df.loc[:, 'quz'::3]
# quz sat
# attempt from 'sat' to 'bar'
df.loc[:, 'sat':'bar']
# no columns returned
# slice from 'sat' to 'bar'
df.loc[:, 'sat':'bar':-1]
sat cat ant quz bar
# slice notation is syntatic sugar for the slice function
# slice from 'quz' to the end by 2 with slice function
df.loc[:, slice('quz',None, 2)]
# quz cat dat
# select specific columns with a list
# select columns foo, bar and dat
df.loc[:, ['foo','bar','dat']]
# foo bar dat
You can slice by rows and columns. For instance, if you have 5 rows with labels v, w, x, y, z
# slice from 'w' to 'y' and 'foo' to 'ant' by 3
df.loc['w':'y', 'foo':'ant':3]
# foo ant
# w
# x
# y
Note: .ix has been deprecated since Pandas v0.20. You should instead use .loc or .iloc, as appropriate.
The DataFrame.ix index is what you want to be accessing. It's a little confusing (I agree that Pandas indexing is perplexing at times!), but the following seems to do what you want:
>>> df = DataFrame(np.random.rand(4,5), columns = list('abcde'))
>>> df.ix[:,'b':]
b c d e
0 0.418762 0.042369 0.869203 0.972314
1 0.991058 0.510228 0.594784 0.534366
2 0.407472 0.259811 0.396664 0.894202
3 0.726168 0.139531 0.324932 0.906575
where .ix[row slice, column slice] is what is being interpreted. More on Pandas indexing here: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-advanced
Lets use the titanic dataset from the seaborn package as an example
# Load dataset (pip install seaborn)
>> import seaborn.apionly as sns
>> titanic = sns.load_dataset('titanic')
using the column names
>> titanic.loc[:,['sex','age','fare']]
using the column indices
>> titanic.iloc[:,[2,3,6]]
using ix (Older than Pandas <.20 version)
>> titanic.ix[:,[‘sex’,’age’,’fare’]]
or
>> titanic.ix[:,[2,3,6]]
using the reindex method
>> titanic.reindex(columns=['sex','age','fare'])
Also, Given a DataFrame
data
as in your example, if you would like to extract column a and d only (e.i. the 1st and the 4th column), iloc mothod from the pandas dataframe is what you need and could be used very effectively. All you need to know is the index of the columns you would like to extract. For example:
>>> data.iloc[:,[0,3]]
will give you
a d
0 0.883283 0.100975
1 0.614313 0.221731
2 0.438963 0.224361
3 0.466078 0.703347
4 0.955285 0.114033
5 0.268443 0.416996
6 0.613241 0.327548
7 0.370784 0.359159
8 0.692708 0.659410
9 0.806624 0.875476
You can slice along the columns of a DataFrame by referring to the names of each column in a list, like so:
data = pandas.DataFrame(np.random.rand(10,5), columns = list('abcde'))
data_ab = data[list('ab')]
data_cde = data[list('cde')]
And if you came here looking for slicing two ranges of columns and combining them together (like me) you can do something like
op = df[list(df.columns[0:899]) + list(df.columns[3593:])]
print op
This will create a new dataframe with first 900 columns and (all) columns > 3593 (assuming you have some 4000 columns in your data set).
Here's how you could use different methods to do selective column slicing, including selective label based, index based and the selective ranges based column slicing.
In [37]: import pandas as pd
In [38]: import numpy as np
In [43]: df = pd.DataFrame(np.random.rand(4,7), columns = list('abcdefg'))
In [44]: df
Out[44]:
a b c d e f g
0 0.409038 0.745497 0.890767 0.945890 0.014655 0.458070 0.786633
1 0.570642 0.181552 0.794599 0.036340 0.907011 0.655237 0.735268
2 0.568440 0.501638 0.186635 0.441445 0.703312 0.187447 0.604305
3 0.679125 0.642817 0.697628 0.391686 0.698381 0.936899 0.101806
In [45]: df.loc[:, ["a", "b", "c"]] ## label based selective column slicing
Out[45]:
a b c
0 0.409038 0.745497 0.890767
1 0.570642 0.181552 0.794599
2 0.568440 0.501638 0.186635
3 0.679125 0.642817 0.697628
In [46]: df.loc[:, "a":"c"] ## label based column ranges slicing
Out[46]:
a b c
0 0.409038 0.745497 0.890767
1 0.570642 0.181552 0.794599
2 0.568440 0.501638 0.186635
3 0.679125 0.642817 0.697628
In [47]: df.iloc[:, 0:3] ## index based column ranges slicing
Out[47]:
a b c
0 0.409038 0.745497 0.890767
1 0.570642 0.181552 0.794599
2 0.568440 0.501638 0.186635
3 0.679125 0.642817 0.697628
### with 2 different column ranges, index based slicing:
In [49]: df[df.columns[0:1].tolist() + df.columns[1:3].tolist()]
Out[49]:
a b c
0 0.409038 0.745497 0.890767
1 0.570642 0.181552 0.794599
2 0.568440 0.501638 0.186635
3 0.679125 0.642817 0.697628
Another way to get a subset of columns from your DataFrame, assuming you want all the rows, would be to do:
data[['a','b']] and data[['c','d','e']]
If you want to use numerical column indexes you can do:
data[data.columns[:2]] and data[data.columns[2:]]
Its equivalent
>>> print(df2.loc[140:160,['Relevance','Title']])
>>> print(df2.ix[140:160,[3,7]])
if Data frame look like that:
group name count
fruit apple 90
fruit banana 150
fruit orange 130
vegetable broccoli 80
vegetable kale 70
vegetable lettuce 125
and OUTPUT could be like
group name count
0 fruit apple 90
1 fruit banana 150
2 fruit orange 130
if you use logical operator np.logical_not
df[np.logical_not(df['group'] == 'vegetable')]
more about
https://docs.scipy.org/doc/numpy-1.13.0/reference/routines.logic.html
other logical operators
logical_and(x1, x2, /[, out, where, ...]) Compute the truth value of
x1 AND x2 element-wise.
logical_or(x1, x2, /[, out, where, casting,
...]) Compute the truth value of x1 OR x2 element-wise.
logical_not(x, /[, out, where, casting, ...]) Compute the truth
value of NOT x element-wise.
logical_xor(x1, x2, /[, out, where, ..]) Compute the truth value of x1 XOR x2, element-wise.
You can use the method truncate
df = pd.DataFrame(np.random.rand(10, 5), columns = list('abcde'))
df_ab = df.truncate(before='a', after='b', axis=1)
df_cde = df.truncate(before='c', axis=1)
How to remove the duplicates in the df? df only has 1 column. In this case "60,25" and "25,60" is a pair of duplicated rows. The output should be the new df. For each pair of duplicated row, the kept row in format "A,B" where A < B, the removed row should be the one A>B. In this case, "25,60" and "80,123" should be kept. For unique row, it should stay whatever it is.
IIUC, using get_dummies with duplicated
df[~df.A.str.get_dummies(sep=',').duplicated()]
Out[956]:
A
0 A,C
1 A,B
4 X,Y,Z
Data input
df
Out[957]:
A
0 A,C
1 A,B
2 C,A
3 B,A
4 X,Y,Z
5 Z,Y,X
Update op change the question totally to different question
newdf=df.A.str.get_dummies(sep=',')
newdf[~newdf.duplicated()].dot(newdf.columns+',').str[:-1]
Out[976]:
0 25,60
1 123,37
dtype: object
I'd do a combination of things.
Use pandas.Series.str.split to split by commas
Use apply(frozenset) to get a hashable set such that I can use duplicated
Use pandas.Series.duplicated with keep='last'
df[~df.A.str.split(',').apply(frozenset).duplicated(keep='last')]
A
1 123,17
3 80,123
4 25,60
5 25,42
Addressing comments
df.A.apply(
lambda x: tuple(sorted(map(int, x.split(','))))
).drop_duplicates().apply(
lambda x: ','.join(map(str, x))
)
0 25,60
1 17,123
2 80,123
5 25,42
Name: A, dtype: object
Setup
df = pd.DataFrame(dict(
A='60,25 123,17 123,80 80,123 25,60 25,42'.split()
))
Can someone explain how these two methods of slicing are different?
I've seen the docs,
and I've seen these answers, but I still find myself unable to understand how the three are different. To me, they seem interchangeable in large part, because they are at the lower levels of slicing.
For example, say we want to get the first five rows of a DataFrame. How is it that these two work?
df.loc[:5]
df.iloc[:5]
Can someone present three cases where the distinction in uses are clearer?
Once upon a time, I also wanted to know how these two functions differ from df.ix[:5] but ix has been removed from pandas 1.0, so I don't care anymore.
Label vs. Location
The main distinction between the two methods is:
loc gets rows (and/or columns) with particular labels.
iloc gets rows (and/or columns) at integer locations.
To demonstrate, consider a series s of characters with a non-monotonic integer index:
>>> s = pd.Series(list("abcdef"), index=[49, 48, 47, 0, 1, 2])
49 a
48 b
47 c
0 d
1 e
2 f
>>> s.loc[0] # value at index label 0
'd'
>>> s.iloc[0] # value at index location 0
'a'
>>> s.loc[0:1] # rows at index labels between 0 and 1 (inclusive)
0 d
1 e
>>> s.iloc[0:1] # rows at index location between 0 and 1 (exclusive)
49 a
Here are some of the differences/similarities between s.loc and s.iloc when passed various objects:
<object>
description
s.loc[<object>]
s.iloc[<object>]
0
single item
Value at index label 0 (the string 'd')
Value at index location 0 (the string 'a')
0:1
slice
Two rows (labels 0 and 1)
One row (first row at location 0)
1:47
slice with out-of-bounds end
Zero rows (empty Series)
Five rows (location 1 onwards)
1:47:-1
slice with negative step
three rows (labels 1 back to 47)
Zero rows (empty Series)
[2, 0]
integer list
Two rows with given labels
Two rows with given locations
s > 'e'
Bool series (indicating which values have the property)
One row (containing 'f')
NotImplementedError
(s>'e').values
Bool array
One row (containing 'f')
Same as loc
999
int object not in index
KeyError
IndexError (out of bounds)
-1
int object not in index
KeyError
Returns last value in s
lambda x: x.index[3]
callable applied to series (here returning 3rd item in index)
s.loc[s.index[3]]
s.iloc[s.index[3]]
loc's label-querying capabilities extend well-beyond integer indexes and it's worth highlighting a couple of additional examples.
Here's a Series where the index contains string objects:
>>> s2 = pd.Series(s.index, index=s.values)
>>> s2
a 49
b 48
c 47
d 0
e 1
f 2
Since loc is label-based, it can fetch the first value in the Series using s2.loc['a']. It can also slice with non-integer objects:
>>> s2.loc['c':'e'] # all rows lying between 'c' and 'e' (inclusive)
c 47
d 0
e 1
For DateTime indexes, we don't need to pass the exact date/time to fetch by label. For example:
>>> s3 = pd.Series(list('abcde'), pd.date_range('now', periods=5, freq='M'))
>>> s3
2021-01-31 16:41:31.879768 a
2021-02-28 16:41:31.879768 b
2021-03-31 16:41:31.879768 c
2021-04-30 16:41:31.879768 d
2021-05-31 16:41:31.879768 e
Then to fetch the row(s) for March/April 2021 we only need:
>>> s3.loc['2021-03':'2021-04']
2021-03-31 17:04:30.742316 c
2021-04-30 17:04:30.742316 d
Rows and Columns
loc and iloc work the same way with DataFrames as they do with Series. It's useful to note that both methods can address columns and rows together.
When given a tuple, the first element is used to index the rows and, if it exists, the second element is used to index the columns.
Consider the DataFrame defined below:
>>> import numpy as np
>>> df = pd.DataFrame(np.arange(25).reshape(5, 5),
index=list('abcde'),
columns=['x','y','z', 8, 9])
>>> df
x y z 8 9
a 0 1 2 3 4
b 5 6 7 8 9
c 10 11 12 13 14
d 15 16 17 18 19
e 20 21 22 23 24
Then for example:
>>> df.loc['c': , :'z'] # rows 'c' and onwards AND columns up to 'z'
x y z
c 10 11 12
d 15 16 17
e 20 21 22
>>> df.iloc[:, 3] # all rows, but only the column at index location 3
a 3
b 8
c 13
d 18
e 23
Sometimes we want to mix label and positional indexing methods for the rows and columns, somehow combining the capabilities of loc and iloc.
For example, consider the following DataFrame. How best to slice the rows up to and including 'c' and take the first four columns?
>>> import numpy as np
>>> df = pd.DataFrame(np.arange(25).reshape(5, 5),
index=list('abcde'),
columns=['x','y','z', 8, 9])
>>> df
x y z 8 9
a 0 1 2 3 4
b 5 6 7 8 9
c 10 11 12 13 14
d 15 16 17 18 19
e 20 21 22 23 24
We can achieve this result using iloc and the help of another method:
>>> df.iloc[:df.index.get_loc('c') + 1, :4]
x y z 8
a 0 1 2 3
b 5 6 7 8
c 10 11 12 13
get_loc() is an index method meaning "get the position of the label in this index". Note that since slicing with iloc is exclusive of its endpoint, we must add 1 to this value if we want row 'c' as well.
iloc works based on integer positioning. So no matter what your row labels are, you can always, e.g., get the first row by doing
df.iloc[0]
or the last five rows by doing
df.iloc[-5:]
You can also use it on the columns. This retrieves the 3rd column:
df.iloc[:, 2] # the : in the first position indicates all rows
You can combine them to get intersections of rows and columns:
df.iloc[:3, :3] # The upper-left 3 X 3 entries (assuming df has 3+ rows and columns)
On the other hand, .loc use named indices. Let's set up a data frame with strings as row and column labels:
df = pd.DataFrame(index=['a', 'b', 'c'], columns=['time', 'date', 'name'])
Then we can get the first row by
df.loc['a'] # equivalent to df.iloc[0]
and the second two rows of the 'date' column by
df.loc['b':, 'date'] # equivalent to df.iloc[1:, 1]
and so on. Now, it's probably worth pointing out that the default row and column indices for a DataFrame are integers from 0 and in this case iloc and loc would work in the same way. This is why your three examples are equivalent. If you had a non-numeric index such as strings or datetimes, df.loc[:5] would raise an error.
Also, you can do column retrieval just by using the data frame's __getitem__:
df['time'] # equivalent to df.loc[:, 'time']
Now suppose you want to mix position and named indexing, that is, indexing using names on rows and positions on columns (to clarify, I mean select from our data frame, rather than creating a data frame with strings in the row index and integers in the column index). This is where .ix comes in:
df.ix[:2, 'time'] # the first two rows of the 'time' column
I think it's also worth mentioning that you can pass boolean vectors to the loc method as well. For example:
b = [True, False, True]
df.loc[b]
Will return the 1st and 3rd rows of df. This is equivalent to df[b] for selection, but it can also be used for assigning via boolean vectors:
df.loc[b, 'name'] = 'Mary', 'John'
In my opinion, the accepted answer is confusing, since it uses a DataFrame with only missing values. I also do not like the term position-based for .iloc and instead, prefer integer location as it is much more descriptive and exactly what .iloc stands for. The key word is INTEGER - .iloc needs INTEGERS.
See my extremely detailed blog series on subset selection for more
.ix is deprecated and ambiguous and should never be used
Because .ix is deprecated we will only focus on the differences between .loc and .iloc.
Before we talk about the differences, it is important to understand that DataFrames have labels that help identify each column and each index. Let's take a look at a sample DataFrame:
df = pd.DataFrame({'age':[30, 2, 12, 4, 32, 33, 69],
'color':['blue', 'green', 'red', 'white', 'gray', 'black', 'red'],
'food':['Steak', 'Lamb', 'Mango', 'Apple', 'Cheese', 'Melon', 'Beans'],
'height':[165, 70, 120, 80, 180, 172, 150],
'score':[4.6, 8.3, 9.0, 3.3, 1.8, 9.5, 2.2],
'state':['NY', 'TX', 'FL', 'AL', 'AK', 'TX', 'TX']
},
index=['Jane', 'Nick', 'Aaron', 'Penelope', 'Dean', 'Christina', 'Cornelia'])
All the words in bold are the labels. The labels, age, color, food, height, score and state are used for the columns. The other labels, Jane, Nick, Aaron, Penelope, Dean, Christina, Cornelia are used for the index.
The primary ways to select particular rows in a DataFrame are with the .loc and .iloc indexers. Each of these indexers can also be used to simultaneously select columns but it is easier to just focus on rows for now. Also, each of the indexers use a set of brackets that immediately follow their name to make their selections.
.loc selects data only by labels
We will first talk about the .loc indexer which only selects data by the index or column labels. In our sample DataFrame, we have provided meaningful names as values for the index. Many DataFrames will not have any meaningful names and will instead, default to just the integers from 0 to n-1, where n is the length of the DataFrame.
There are three different inputs you can use for .loc
A string
A list of strings
Slice notation using strings as the start and stop values
Selecting a single row with .loc with a string
To select a single row of data, place the index label inside of the brackets following .loc.
df.loc['Penelope']
This returns the row of data as a Series
age 4
color white
food Apple
height 80
score 3.3
state AL
Name: Penelope, dtype: object
Selecting multiple rows with .loc with a list of strings
df.loc[['Cornelia', 'Jane', 'Dean']]
This returns a DataFrame with the rows in the order specified in the list:
Selecting multiple rows with .loc with slice notation
Slice notation is defined by a start, stop and step values. When slicing by label, pandas includes the stop value in the return. The following slices from Aaron to Dean, inclusive. Its step size is not explicitly defined but defaulted to 1.
df.loc['Aaron':'Dean']
Complex slices can be taken in the same manner as Python lists.
.iloc selects data only by integer location
Let's now turn to .iloc. Every row and column of data in a DataFrame has an integer location that defines it. This is in addition to the label that is visually displayed in the output. The integer location is simply the number of rows/columns from the top/left beginning at 0.
There are three different inputs you can use for .iloc
An integer
A list of integers
Slice notation using integers as the start and stop values
Selecting a single row with .iloc with an integer
df.iloc[4]
This returns the 5th row (integer location 4) as a Series
age 32
color gray
food Cheese
height 180
score 1.8
state AK
Name: Dean, dtype: object
Selecting multiple rows with .iloc with a list of integers
df.iloc[[2, -2]]
This returns a DataFrame of the third and second to last rows:
Selecting multiple rows with .iloc with slice notation
df.iloc[:5:3]
Simultaneous selection of rows and columns with .loc and .iloc
One excellent ability of both .loc/.iloc is their ability to select both rows and columns simultaneously. In the examples above, all the columns were returned from each selection. We can choose columns with the same types of inputs as we do for rows. We simply need to separate the row and column selection with a comma.
For example, we can select rows Jane, and Dean with just the columns height, score and state like this:
df.loc[['Jane', 'Dean'], 'height':]
This uses a list of labels for the rows and slice notation for the columns
We can naturally do similar operations with .iloc using only integers.
df.iloc[[1,4], 2]
Nick Lamb
Dean Cheese
Name: food, dtype: object
Simultaneous selection with labels and integer location
.ix was used to make selections simultaneously with labels and integer location which was useful but confusing and ambiguous at times and thankfully it has been deprecated. In the event that you need to make a selection with a mix of labels and integer locations, you will have to make both your selections labels or integer locations.
For instance, if we want to select rows Nick and Cornelia along with columns 2 and 4, we could use .loc by converting the integers to labels with the following:
col_names = df.columns[[2, 4]]
df.loc[['Nick', 'Cornelia'], col_names]
Or alternatively, convert the index labels to integers with the get_loc index method.
labels = ['Nick', 'Cornelia']
index_ints = [df.index.get_loc(label) for label in labels]
df.iloc[index_ints, [2, 4]]
Boolean Selection
The .loc indexer can also do boolean selection. For instance, if we are interested in finding all the rows wher age is above 30 and return just the food and score columns we can do the following:
df.loc[df['age'] > 30, ['food', 'score']]
You can replicate this with .iloc but you cannot pass it a boolean series. You must convert the boolean Series into a numpy array like this:
df.iloc[(df['age'] > 30).values, [2, 4]]
Selecting all rows
It is possible to use .loc/.iloc for just column selection. You can select all the rows by using a colon like this:
df.loc[:, 'color':'score':2]
The indexing operator, [], can select rows and columns too but not simultaneously.
Most people are familiar with the primary purpose of the DataFrame indexing operator, which is to select columns. A string selects a single column as a Series and a list of strings selects multiple columns as a DataFrame.
df['food']
Jane Steak
Nick Lamb
Aaron Mango
Penelope Apple
Dean Cheese
Christina Melon
Cornelia Beans
Name: food, dtype: object
Using a list selects multiple columns
df[['food', 'score']]
What people are less familiar with, is that, when slice notation is used, then selection happens by row labels or by integer location. This is very confusing and something that I almost never use but it does work.
df['Penelope':'Christina'] # slice rows by label
df[2:6:2] # slice rows by integer location
The explicitness of .loc/.iloc for selecting rows is highly preferred. The indexing operator alone is unable to select rows and columns simultaneously.
df[3:5, 'color']
TypeError: unhashable type: 'slice'
.loc and .iloc are used for indexing, i.e., to pull out portions of data. In essence, the difference is that .loc allows label-based indexing, while .iloc allows position-based indexing.
If you get confused by .loc and .iloc, keep in mind that .iloc is based on the index (starting with i) position, while .loc is based on the label (starting with l).
.loc
.loc is supposed to be based on the index labels and not the positions, so it is analogous to Python dictionary-based indexing. However, it can accept boolean arrays, slices, and a list of labels (none of which work with a Python dictionary).
iloc
.iloc does the lookup based on index position, i.e., pandas behaves similarly to a Python list. pandas will raise an IndexError if there is no index at that location.
Examples
The following examples are presented to illustrate the differences between .iloc and .loc. Let's consider the following series:
>>> s = pd.Series([11, 9], index=["1990", "1993"], name="Magic Numbers")
>>> s
1990 11
1993 9
Name: Magic Numbers , dtype: int64
.iloc Examples
>>> s.iloc[0]
11
>>> s.iloc[-1]
9
>>> s.iloc[4]
Traceback (most recent call last):
...
IndexError: single positional indexer is out-of-bounds
>>> s.iloc[0:3] # slice
1990 11
1993 9
Name: Magic Numbers , dtype: int64
>>> s.iloc[[0,1]] # list
1990 11
1993 9
Name: Magic Numbers , dtype: int64
.loc Examples
>>> s.loc['1990']
11
>>> s.loc['1970']
Traceback (most recent call last):
...
KeyError: ’the label [1970] is not in the [index]’
>>> mask = s > 9
>>> s.loc[mask]
1990 11
Name: Magic Numbers , dtype: int64
>>> s.loc['1990':] # slice
1990 11
1993 9
Name: Magic Numbers, dtype: int64
Because s has string index values, .loc will fail when
indexing with an integer:
>>> s.loc[0]
Traceback (most recent call last):
...
KeyError: 0
This example will illustrate the difference:
df = pd.DataFrame({'col1': [1,2,3,4,5], 'col2': ["foo", "bar", "baz", "foobar", "foobaz"]})
col1 col2
0 1 foo
1 2 bar
2 3 baz
3 4 foobar
4 5 foobaz
df = df.sort_values('col1', ascending = False)
col1 col2
4 5 foobaz
3 4 foobar
2 3 baz
1 2 bar
0 1 foo
Index based access:
df.iloc[0, 0:2]
col1 5
col2 foobaz
Name: 4, dtype: object
We get the first row of the sorted dataframe. (This is not the row with index 0, but with index 4).
Position based access:
df.loc[0, 'col1':'col2']
col1 1
col2 foo
Name: 0, dtype: object
We get the row with index 0, even when the df is sorted.
DataFrame.loc() : Select rows by index value
DataFrame.iloc() : Select rows by rows number
Example:
Select first 5 rows of a table, df1 is your dataframe
df1.iloc[:5]
Select first A, B rows of a table, df1 is your dataframe
df1.loc['A','B']