I'm currently working on some python dataframes over on pandas. And I'm not sure how this operation can be done. For example, I have an empty dataframe df and list of the following triples:
L = [(1,2,3), (2,5,4), (2,5,4), (3,2,0), (2,1,3)]
I wish to add all these triples into the dataframe df with columns ['id', 'a', 'b', 'c'] according to some constraint. The id is simply a counter that determines how many items have been added so far and a, b, and c are columns for the triples (but they would be commutative with each other). So the idea is to linearly traverse all items in L and then add each one to the df according to the restriction:
It is ok to add (1,2,3) since df is still empty. (id=0)
It is ok to add (2,5,4) since it or any of its permutation has not appeared yet in df. (id=1)
We then see (2,5,4) but this already exists in df, hence we cannot add it.
Next is (3,2,0) and we can clearly add this for the same reason as #2. (id=2)
Finally, it's (2,1,3). While this triple has not existed yet in df but since it's a permutation to an existing triplet in df (which is the (1,2,3)), then we cannot add it to df.
In the end, the final df should look something like this.
id a b c
0 1 2 3
1 2 5 4
2 3 2 0
Anyone knows how this can be done? My idea is to first curate an auxiliary list LL that would contain these "unique" triples and then just transform it into a pandas df. But I'm not sure if it's a fast and elegant efficient approach.
Fast solution
Create a numpy array from the list, then sort the array along axis=1 and use duplicated to create a boolean mask to identify dupes, then remove the duplicate rows from the array and create a new dataframe
a = np.array(L)
m = pd.DataFrame(np.sort(a, axis=1)).duplicated()
pd.DataFrame(a[~m], columns=['a', 'b', 'c'])
Result
a b c
0 1 2 3
1 2 5 4
2 3 2 0
You can use a dictionary comprehension with a frozenset of the tuple as key to eliminate the duplicated permutations, then feed the values to the DataFrame constructor:
L = [(1,2,3), (2,5,4), (2,5,4), (3,2,0), (2,1,3)]
df = pd.DataFrame({frozenset(t): t for t in L[::-1]}.values(),
columns=['a', 'b', 'c'])
output:
a b c
0 1 2 3
1 3 2 0
2 2 5 4
If order is important, you can use a set to collect the seen values instead:
seen = set()
df = pd.DataFrame([t for t in L if (f:=frozenset(t)) not in seen
and not seen.add(f)],
columns=['a', 'b', 'c'])
output:
a b c
0 1 2 3
1 2 5 4
2 3 2 0
handling duplicates values in the tuple
df = pd.DataFrame({tuple(sorted(t)): t
for t in L[::-1]}.values(),
columns=['a', 'b', 'c'])
If there are many columns, sorting becomes inefficient, then you can use a Counter:
from collections import Counter
df = pd.DataFrame({frozenset(Counter(t).items()): t
for t in L[::-1]}.values(),
columns=['a', 'b', 'c'])
pure pandas alternative:
You can do the same with pandas using loc and aggregation to set:
df = pd.DataFrame(L).loc[lambda d: ~d.agg(set, axis=1).duplicated()]
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))
The pandas drop_duplicates function is great for "uniquifying" a dataframe. I would like to drop all rows which are duplicates across a subset of columns. Is this possible?
A B C
0 foo 0 A
1 foo 1 A
2 foo 1 B
3 bar 1 A
As an example, I would like to drop rows which match on columns A and C so this should drop rows 0 and 1.
This is much easier in pandas now with drop_duplicates and the keep parameter.
import pandas as pd
df = pd.DataFrame({"A":["foo", "foo", "foo", "bar"], "B":[0,1,1,1], "C":["A","A","B","A"]})
df.drop_duplicates(subset=['A', 'C'], keep=False)
Just want to add to Ben's answer on drop_duplicates:
keep : {‘first’, ‘last’, False}, default ‘first’
first : Drop duplicates except for the first occurrence.
last : Drop duplicates except for the last occurrence.
False : Drop all duplicates.
So setting keep to False will give you desired answer.
DataFrame.drop_duplicates(*args, **kwargs) Return DataFrame with
duplicate rows removed, optionally only considering certain columns
Parameters: subset : column label or sequence of labels, optional
Only consider certain columns for identifying duplicates, by default
use all of the columns keep : {‘first’, ‘last’, False}, default
‘first’ first : Drop duplicates except for the first occurrence. last
: Drop duplicates except for the last occurrence. False : Drop all
duplicates. take_last : deprecated inplace : boolean, default False
Whether to drop duplicates in place or to return a copy cols : kwargs
only argument of subset [deprecated] Returns: deduplicated :
DataFrame
If you want result to be stored in another dataset:
df.drop_duplicates(keep=False)
or
df.drop_duplicates(keep=False, inplace=False)
If same dataset needs to be updated:
df.drop_duplicates(keep=False, inplace=True)
Above examples will remove all duplicates and keep one, similar to DISTINCT * in SQL
use groupby and filter
import pandas as pd
df = pd.DataFrame({"A":["foo", "foo", "foo", "bar"], "B":[0,1,1,1], "C":["A","A","B","A"]})
df.groupby(["A", "C"]).filter(lambda df:df.shape[0] == 1)
Try these various things
df = pd.DataFrame({"A":["foo", "foo", "foo", "bar","foo"], "B":[0,1,1,1,1], "C":["A","A","B","A","A"]})
>>>df.drop_duplicates( "A" , keep='first')
or
>>>df.drop_duplicates( keep='first')
or
>>>df.drop_duplicates( keep='last')
Actually, drop rows 0 and 1 only requires (any observations containing matched A and C is kept.):
In [335]:
df['AC']=df.A+df.C
In [336]:
print df.drop_duplicates('C', take_last=True) #this dataset is a special case, in general, one may need to first drop_duplicates by 'c' and then by 'a'.
A B C AC
2 foo 1 B fooB
3 bar 1 A barA
[2 rows x 4 columns]
But I suspect what you really want is this (one observation containing matched A and C is kept.):
In [337]:
print df.drop_duplicates('AC')
A B C AC
0 foo 0 A fooA
2 foo 1 B fooB
3 bar 1 A barA
[3 rows x 4 columns]
Edit:
Now it is much clearer, therefore:
In [352]:
DG=df.groupby(['A', 'C'])
print pd.concat([DG.get_group(item) for item, value in DG.groups.items() if len(value)==1])
A B C
2 foo 1 B
3 bar 1 A
[2 rows x 3 columns]
You can use duplicated() to flag all duplicates and filter out flagged rows. If you need to assign columns to new_df later, make sure to call .copy() so that you don't get SettingWithCopyWarning later on.
new_df = df[~df.duplicated(subset=['A', 'C'], keep=False)].copy()
One nice feature of this method is that you can conditionally drop duplicates with it. For example, to drop all duplicated rows only if column A is equal to 'foo', you can use the following code.
new_df = df[~( df.duplicated(subset=['A', 'B', 'C'], keep=False) & df['A'].eq('foo') )].copy()
Also, if you don't wish to write out columns by name, you can pass slices of df.columns to subset=. This is also true for drop_duplicates() as well.
# to consider all columns for identifying duplicates
df[~df.duplicated(subset=df.columns, keep=False)].copy()
# the same is true for drop_duplicates
df.drop_duplicates(subset=df.columns, keep=False)
# to consider columns in positions 0 and 2 (i.e. 'A' and 'C') for identifying duplicates
df.drop_duplicates(subset=df.columns[[0, 2]], keep=False)
If you want to check 2 columns with try and except statements, this one can help out.
if "column_2" in df.columns:
try:
df[['column_1', "column_2"]] = df[['header', "column_2"]].drop_duplicates(subset = ["column_2", "column_1"] ,keep="first")
except:
df[["column_2"]] = df[["column_2"]].drop_duplicates(subset="column_2" ,keep="first")
print(f"No column_1 for {path}.")
try:
df[["column_1"]] = df[["column_1"]].drop_duplicates(subset="column_1" ,keep="first")
except:
print(f"No column_1 or column_2 for {path}.")
I am using .size() on a groupby result in order to count how many items are in each group.
I would like the result to be saved to a new column name without manually editing the column names array, how can it be done?
This is what I have tried:
grpd = df.groupby(['A','B'])
grpd['size'] = grpd.size()
grpd
and the error I got:
TypeError: 'DataFrameGroupBy' object does not support item assignment
(on the second line)
The .size() built-in method of DataFrameGroupBy objects actually returns a Series object with the group sizes and not a DataFrame. If you want a DataFrame whose column is the group sizes, indexed by the groups, with a custom name, you can use the .to_frame() method and use the desired column name as its argument.
grpd = df.groupby(['A','B']).size().to_frame('size')
If you wanted the groups to be columns again you could add a .reset_index() at the end.
You need transform size - len of df is same as before:
Notice:
Here it is necessary to add one column after groupby, else you get an error. Because GroupBy.size count NaNs too, what column is used is not important. All columns working same.
import pandas as pd
df = pd.DataFrame({'A': ['x', 'x', 'x','y','y']
, 'B': ['a', 'c', 'c','b','b']})
print (df)
A B
0 x a
1 x c
2 x c
3 y b
4 y b
df['size'] = df.groupby(['A', 'B'])['A'].transform('size')
print (df)
A B size
0 x a 1
1 x c 2
2 x c 2
3 y b 2
4 y b 2
If need set column name in aggregating df - len of df is obviously NOT same as before:
import pandas as pd
df = pd.DataFrame({'A': ['x', 'x', 'x','y','y']
, 'B': ['a', 'c', 'c','b','b']})
print (df)
A B
0 x a
1 x c
2 x c
3 y b
4 y b
df = df.groupby(['A', 'B']).size().reset_index(name='Size')
print (df)
A B Size
0 x a 1
1 x c 2
2 y b 2
The result of df.groupby(...) is not a DataFrame. To get a DataFrame back, you have to apply a function to each group, transform each element of a group, or filter the groups.
It seems like you want a DataFrame that contains (1) all your original data in df and (2) the count of how much data is in each group. These things have different lengths, so if they need to go into the same DataFrame, you'll need to list the size redundantly, i.e., for each row in each group.
df['size'] = df.groupby(['A','B']).transform(np.size)
(Aside: It's helpful if you can show succinct sample input and expected results.)
You can set the as_index parameter in groupby to False to get a DataFrame instead of a Series:
df = pd.DataFrame({'A': ['a', 'a', 'b', 'b'], 'B': [1, 2, 2, 2]})
df.groupby(['A', 'B'], as_index=False).size()
Output:
A B size
0 a 1 1
1 a 2 1
2 b 2 2
lets say n is the name of dataframe and cst is the no of items being repeted.
Below code gives the count in next column
cstn=Counter(n.cst)
cstlist = pd.DataFrame.from_dict(cstn, orient='index').reset_index()
cstlist.columns=['name','cnt']
n['cnt']=n['cst'].map(cstlist.loc[:, ['name','cnt']].set_index('name').iloc[:,0].to_dict())
Hope this will work
So far I am able to get the list of all column names present in the dataframe or to get a specific column names based on its datatype, starting letters, etc...
Now my requirement is to get the whole list of column names or a sublist and to exclude one column from it (i.e Target variable / Label Column. This is a part of Machine Learning. So I am using the terms that are used in machine learning)
Please note I am not speaking about the data present in those columns. I am just taking the column names and want to exclude a particular column by its name
Please see below example for better understanding :
# Get all the column names from a Dataframe
df.columns
Index(['transactionID', 'accountID', 'transactionAmountUSD',
'transactionAmount', 'transactionCurrencyCode',
'accountAge', 'validationid', 'LABEL'],
dtype='object')
# Get only the Numeric Variables (Columns with numeric values in it)
df._get_numeric_data().columns
Index(['transactionAmountUSD', 'transactionAmount', 'accountAge', 'LABEL'],
dtype='object')
Now inorder to get remaining column names I am subtracting both the above commands
string_cols = list(set(list(df.columns))-set(df._get_numeric_data().columns))
Ok everything goes well until I hit this.
I have found out that Label column though it has numeric values it should not be present in the list of numeric variables. It should be excluded.
(i.e) I want to exclude a particular column name (not using its index in the list but using its name explicitly)
I tried similar statements like the following ones but in vain. Any inputs on this will be helpful
set(df._get_numeric_data().columns-set(df.LABEL)
set(df._get_numeric_data().columns-set(df.LABEL.column)
set(df._get_numeric_data().columns-set(df['LABEL'])
I am sure I am missing a very basic thing but not able to figure it out.
First of all, you can exclude all numeric columns much more simply with
pd.DataFrame.select_dtypes(exclude=[np.number])
transactionID accountID transactionCurrencyCode validationid
0 a a a a
1 a a a a
2 a a a a
3 a a a a
4 a a a a
Second of all, there are many ways to drop a column. See this post
df._get_numeric_data().drop('LABEL', 1)
transactionAmountUSD transactionAmount accountAge
0 1 1 1
1 1 1 1
2 1 1 1
3 1 1 1
4 1 1 1
If you really wanted the columns, use pd.Index.difference
df._get_numeric_data().columns.difference(['LABEL'])
Index(['accountAge', 'transactionAmount', 'transactionAmountUSD'], dtype='object')
Setup
df = pd.DataFrame(
[['a', 'a', 1, 1, 'a', 1, 'a', 1]] * 5,
columns=[
'transactionID', 'accountID', 'transactionAmountUSD',
'transactionAmount', 'transactionCurrencyCode',
'accountAge', 'validationid', 'LABEL']
)
Pandas' index supports set operations, so to exclude one column from column index you can just write something like
import pandas as pd
df = pd.DataFrame(columns=list('abcdef'))
print(df.columns.difference({'b'}))
which will return to you
Index(['a', 'c', 'd', 'e', 'f'], dtype='object')
I hope this is what you want :)
Considering LABEL column as your output and the other features as your input, you can try this:
feature_names = [x for x in df._get_numeric_data().columns if x not in ['LABEL']]
input = df[feature_names]
output= df['LABEL']
Hope this helps.