Normalize Column Values by Monthly Averages with added Group dimension - python-3.x

Initial Note
I already got this running, but it takes a very long time to execute. My DataFrame is around 500MB large. I am hoping to hear some feedback on how to execute this as quickly as possible.
Problem Statement
I want to normalize the DataFrame columns by the mean of the column's values during each month. An added complexity is that I have a column named group which denotes a different sensor in which the parameter (column) was measured. Therefore, the analysis needs to iterate around group and each month.
DF example
X Y Z group
2019-02-01 09:30:07 1 2 1 'grp1'
2019-02-01 09:30:23 2 4 3 'grp2'
2019-02-01 09:30:38 3 6 5 'grp1'
...
Code (Functional, but slow)
This is the code that I used. Coding annotations provide descriptions of most lines. I recognize that the three for loops are causing this runtime issue, but I do not have the foresight to see a way around it. Does anyone know any
# Get mean monthly values for each group
mean_per_month_unit = process_df.groupby('group').resample('M', how='mean')
# Store the monthly dates created in last line into a list called month_dates
month_dates = mean_per_month_unit.index.get_level_values(1)
# Place date on multiIndex columns. future note: use df[DATE, COL_NAME][UNIT] to access mean value
mean_per_month_unit = mean_per_month_unit.unstack().swaplevel(0,1,1).sort_index(axis=1)
divide_df = pd.DataFrame().reindex_like(df)
process_cols.remove('group')
for grp in group_list:
print(grp)
# Iterate through month
for mnth in month_dates:
# Make mask where month and group
mask = (df.index.month == mnth.month) & (df['group'] == grp)
for col in process_cols:
# Set values of divide_df
divide_df.iloc[mask.tolist(), divide_df.columns.get_loc(col)] = mean_per_month_unit[mnth, col][grp]
# Divide process_df with divide_df
final_df = process_df / divide_df.values
EDIT: Example data
Here is the data in CSV format.
EDIT2: Current code (according to current answer)
def normalize_df(df):
df['month'] = df.index.month
print(df['month'])
df['year'] = df.index.year
print(df['year'])
def find_norm(x, df_col_list): # x is a row in dataframe, col_list is the list of columns to normalize
agg = df.groupby(by=['group', 'month', 'year'], as_index=True).mean()
print("###################", x.name, x['month'])
for column in df_col_list: # iterate over col list, find mean from aggregations, and divide the value by
print(column)
mean_col = agg.loc[(x['group'], x['month'], x['year']), column]
print(mean_col)
col_name = "norm" + str(column)
x[col_name] = x[column] / mean_col # norm
return x
normalize_cols = df.columns.tolist()
normalize_cols.remove('group')
#normalize_cols.remove('mode')
df2 = df.apply(find_norm, df_col_list = normalize_cols, axis=1)
The code runs perfectly for one iteration and then it fails with the error:
KeyError: ('month', 'occurred at index 2019-02-01 11:30:17')
As I said, it runs correctly once. However, it iterates over the same row again and then fails. I see according to df.apply() documentation that the first row always runs twice. I'm just not sure why this fails on the second time through.

Assuming that the requirement is to group the columns by mean and the month, here is another approach:
Create new columns - month and year from the index. df.index.month can be used for this provided the index is of type DatetimeIndex
type(df.index) # df is the original dataframe
#pandas.core.indexes.datetimes.DatetimeIndex
df['month'] = df.index.month
df['year'] = df.index.year # added year assuming the grouping occurs per grp per month per year. No need to add this column if year is not to be considered.
Now, group over (grp, month, year) and aggregate to find mean of every column. (Added year assuming the grouping occurs per grp per month per year. No need to add this column if year is not to be considered.)
agg = df.groupby(by=['grp', 'month', 'year'], as_index=True).mean()
Use a function to calculate the normalized values and use apply() over the original dataframe
def find_norm(x, df_col_list): # x is a row in dataframe, col_list is the list of columns to normalize
for column in df_col_list: # iterate over col list, find mean from aggregations, and divide the value by the mean.
mean_col = agg.loc[(str(x['grp']), x['month'], x['year']), column]
col_name = "norm" + str(column)
x[col_name] = x[column] / mean_col # norm
return x
df2 = df.apply(find_norm, df_col_list = ['A','B','C'], axis=1)
#df2 will now have 3 additional columns - normA, normB, normC
df2:
A B C grp month year normA normB normC
2019-02-01 09:30:07 1 2 3 1 2 2019 0.666667 0.8 1.5
2019-03-02 09:30:07 2 3 4 1 3 2019 1.000000 1.0 1.0
2019-02-01 09:40:07 2 3 1 2 2 2019 1.000000 1.0 1.0
2019-02-01 09:38:07 2 3 1 1 2 2019 1.333333 1.2 0.5
Alternatively, for step 3, one can join the agg and df dataframes and find the norm.
Hope this helps!
Here is how the code would look like:
# Step 1
df['month'] = df.index.month
df['year'] = df.index.year # added year assuming the grouping occurs
# Step 2
agg = df.groupby(by=['grp', 'month', 'year'], as_index=True).mean()
# Step 3
def find_norm(x, df_col_list): # x is a row in dataframe, col_list is the list of columns to normalize
for column in df_col_list: # iterate over col list, find mean from aggregations, and divide the value by the mean.
mean_col = agg.loc[(str(x['grp']), x['month'], x['year']), column]
col_name = "norm" + str(column)
x[col_name] = x[column] / mean_col # norm
return x
df2 = df.apply(find_norm, df_col_list = ['A','B','C'], axis=1)

Related

Add Column For Results Of Dataframe Resample [duplicate]

I have the following data frame in IPython, where each row is a single stock:
In [261]: bdata
Out[261]:
<class 'pandas.core.frame.DataFrame'>
Int64Index: 21210 entries, 0 to 21209
Data columns:
BloombergTicker 21206 non-null values
Company 21210 non-null values
Country 21210 non-null values
MarketCap 21210 non-null values
PriceReturn 21210 non-null values
SEDOL 21210 non-null values
yearmonth 21210 non-null values
dtypes: float64(2), int64(1), object(4)
I want to apply a groupby operation that computes cap-weighted average return across everything, per each date in the "yearmonth" column.
This works as expected:
In [262]: bdata.groupby("yearmonth").apply(lambda x: (x["PriceReturn"]*x["MarketCap"]/x["MarketCap"].sum()).sum())
Out[262]:
yearmonth
201204 -0.109444
201205 -0.290546
But then I want to sort of "broadcast" these values back to the indices in the original data frame, and save them as constant columns where the dates match.
In [263]: dateGrps = bdata.groupby("yearmonth")
In [264]: dateGrps["MarketReturn"] = dateGrps.apply(lambda x: (x["PriceReturn"]*x["MarketCap"]/x["MarketCap"].sum()).sum())
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
/mnt/bos-devrnd04/usr6/home/espears/ws/Research/Projects/python-util/src/util/<ipython-input-264-4a68c8782426> in <module>()
----> 1 dateGrps["MarketReturn"] = dateGrps.apply(lambda x: (x["PriceReturn"]*x["MarketCap"]/x["MarketCap"].sum()).sum())
TypeError: 'DataFrameGroupBy' object does not support item assignment
I realize this naive assignment should not work. But what is the "right" Pandas idiom for assigning the result of a groupby operation into a new column on the parent dataframe?
In the end, I want a column called "MarketReturn" than will be a repeated constant value for all indices that have matching date with the output of the groupby operation.
One hack to achieve this would be the following:
marketRetsByDate = dateGrps.apply(lambda x: (x["PriceReturn"]*x["MarketCap"]/x["MarketCap"].sum()).sum())
bdata["MarketReturn"] = np.repeat(np.NaN, len(bdata))
for elem in marketRetsByDate.index.values:
bdata["MarketReturn"][bdata["yearmonth"]==elem] = marketRetsByDate.ix[elem]
But this is slow, bad, and unPythonic.
In [97]: df = pandas.DataFrame({'month': np.random.randint(0,11, 100), 'A': np.random.randn(100), 'B': np.random.randn(100)})
In [98]: df.join(df.groupby('month')['A'].sum(), on='month', rsuffix='_r')
Out[98]:
A B month A_r
0 -0.040710 0.182269 0 -0.331816
1 -0.004867 0.642243 1 2.448232
2 -0.162191 0.442338 4 2.045909
3 -0.979875 1.367018 5 -2.736399
4 -1.126198 0.338946 5 -2.736399
5 -0.992209 -1.343258 1 2.448232
6 -1.450310 0.021290 0 -0.331816
7 -0.675345 -1.359915 9 2.722156
While I'm still exploring all of the incredibly smart ways that apply concatenates the pieces it's given, here's another way to add a new column in the parent after a groupby operation.
In [236]: df
Out[236]:
yearmonth return
0 201202 0.922132
1 201202 0.220270
2 201202 0.228856
3 201203 0.277170
4 201203 0.747347
In [237]: def add_mkt_return(grp):
.....: grp['mkt_return'] = grp['return'].sum()
.....: return grp
.....:
In [238]: df.groupby('yearmonth').apply(add_mkt_return)
Out[238]:
yearmonth return mkt_return
0 201202 0.922132 1.371258
1 201202 0.220270 1.371258
2 201202 0.228856 1.371258
3 201203 0.277170 1.024516
4 201203 0.747347 1.024516
As a general rule when using groupby(), if you use the .transform() function pandas will return a table with the same length as your original. When you use other functions like .sum() or .first() then pandas will return a table where each row is a group.
I'm not sure how this works with apply but implementing elaborate lambda functions with transform can be fairly tricky so the strategy that I find most helpful is to create the variables I need, place them in the original dataset and then do my operations there.
If I understand what you're trying to do correctly first you can calculate the total market cap for each group:
bdata['group_MarketCap'] = bdata.groupby('yearmonth')['MarketCap'].transform('sum')
This will add a column called "group_MarketCap" to your original data which would contain the sum of market caps for each group. Then you can calculate the weighted values directly:
bdata['weighted_P'] = bdata['PriceReturn'] * (bdata['MarketCap']/bdata['group_MarketCap'])
And finally you would calculate the weighted average for each group using the same transform function:
bdata['MarketReturn'] = bdata.groupby('yearmonth')['weighted_P'].transform('sum')
I tend to build my variables this way. Sometimes you can pull off putting it all in a single command but that doesn't always work with groupby() because most of the time pandas needs to instantiate the new object to operate on it at the full dataset scale (i.e. you can't add two columns together if one doesn't exist yet).
Hope this helps :)
May I suggest the transform method (instead of aggregate)? If you use it in your original example it should do what you want (the broadcasting).
I did not find a way to make assignment to the original dataframe. So I just store the results from the groups and concatenate them. Then we sort the concatenated dataframe by index to get the original order as the input dataframe. Here is a sample code:
In [10]: df = pd.DataFrame({'month': np.random.randint(0,11, 100), 'A': np.random.randn(100), 'B': np.random.randn(100)})
In [11]: df.head()
Out[11]:
month A B
0 4 -0.029106 -0.904648
1 2 -2.724073 0.492751
2 7 0.732403 0.689530
3 2 0.487685 -1.017337
4 1 1.160858 -0.025232
In [12]: res = []
In [13]: for month, group in df.groupby('month'):
...: new_df = pd.DataFrame({
...: 'A^2+B': group.A ** 2 + group.B,
...: 'A+B^2': group.A + group.B**2
...: })
...: res.append(new_df)
...:
In [14]: res = pd.concat(res).sort_index()
In [15]: res.head()
Out[15]:
A^2+B A+B^2
0 -0.903801 0.789282
1 7.913327 -2.481270
2 1.225944 1.207855
3 -0.779501 1.522660
4 1.322360 1.161495
This method is pretty fast and extensible. You can derive any feature here.
Note: If the dataframe is too large, concat may cause you MMO error.

Merge two Dataframes in combination with .isin() or .contains() or difflib? [duplicate]

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))

Add new rows to dataframe using existing rows from previous year

I'm creating a Pandas dataframe from an existing file and it ends up essentially like this.
import pandas as pd
import datetime
data = [[i, i+1] for i in range(14)]
index = pd.date_range(start=datetime.date(2019,1,1), end=datetime.date(2020,2,1), freq='MS')
columns = ['col1', 'col2']
df = pd.DataFrame(data, index, columns)
Notice that this doesn't go all the way up to the present -- often the file I'm pulling from is a month or two behind. What I then need to do is add on any missing months and fill them with the same value as the previous year.
So in this case I need to add another row that is
2020-03-01 2 3
It could be anywhere from 0-2 rows that need to be added to the end of the dataframe at a given point in time. What's the best way to do this?
Note: The data here is not real so please don't take advantage of the simple pattern of entries I gave above. It was just a quick way to fill two columns of a table as an example.
If I understand your problem, then the following should help you. This does assume that you always have data 12 months ago however. You can define a new DataFrame which includes the months up to the most recent date.
# First create the new index. Get the most recent date and add an offset.
start, end = df.index[-1] + pd.DateOffset(), pd.Timestamp.now()
index_new = pd.date_range(start, end, freq='MS')
Create your DataFrame
# Get the data from the previous year.
data = df.loc[index_new - pd.DateOffset(years=1)].values
df_new = pd.DataFrame(data, index = index_new, columns=df.columns)
which looks like
col1 col2
2020-03-01 2 3
then just use;
pd.concat([df, df_new], axis=0)
Which gives
col1 col2
2019-01-01 0 1
2019-02-01 1 2
2019-03-01 2 3
... ... ...
2020-02-01 13 14
2020-03-01 2 3
Note
This also works for cases where the number of months missing is greater than 1.
Edit
Slightly different variation
# Create series with missing months added.
# Get the corresponding data 12 months prior.
s = pd.date_range(df.index[0], pd.Timestamp.now(), freq='MS')
fill = df.loc[s[~s.isin(df.index)] - pd.DateOffset(years=1)]
# Reindex the original dataframe
df = df.reindex(s)
# Find the dates to fill and replace with lagged data
df.iloc[-1 * fill.shape[0]:] = fill.values

How to return index of a row 60 seconds before current row

I have a large (>32 M rows) Pandas dataframe.
In column 'Time_Stamp' I have a Unix timestamp in seconds. These values are not linear, there are gaps, and some timestamps can be duplicated (ex: 1, 2, 4, 6, 6, 9,...).
I would like to set column 'Result' of current row to the index of the row that is 60 seconds before current row (closest match if there are no rows exactly 60 seconds before current row, and if more than one match, take maximum of all matches).
I've tried this to first get the list of indexes, but it always return an empty list:
df.index[df['Time_Stamp'] <= df.Time_Stamp-60].tolist()
I cannot use a for loop due to the large number of rows.
Edit 20.01.2020:
Based on comment below, I'm adding a sample dataset, and instead of returning the index I want to return the column Value:
In [2]: df
Out[2]:
Time_Stamp Value
0 1 2.4
1 2 3.1
2 4 6.3
3 6 7.2
4 6 6.1
5 9 6.0
So with the precious help of ALollz, I managed to achieve what i wanted to do in the end, here's my code:
#make copy of dataframe
df2 = df[['Time_Stamp','Value']].copy()
#add Time_gap to Time_Stamp in df2
df2['Time_Stamp'] = df2.Time_Stamp +Time_gap
#sort df2 on Time_Stamp
df2.sort_values(by = 'Time_Stamp', ascending=True,inplace = True)
df2 = df2.reset_index(drop=True)
df3 = pd.merge_asof(df, df2, on='Time_Stamp', direction='forward')

How can I count categorical columns by month in Pandas?

I have time series data with a column which can take a value A, B, or C.
An example of my data looks like this:
date,category
2017-01-01,A
2017-01-15,B
2017-01-20,A
2017-02-02,C
2017-02-03,A
2017-02-05,C
2017-02-08,C
I want to group my data by month and store both the sum of the count of A and the count of B in column a_or_b_count and the count of C in c_count.
I've tried several things, but the closest I've been able to do is to preprocess the data with the following function:
def preprocess(df):
# Remove everything more granular than day by splitting the stringified version of the date.
df['date'] = pd.to_datetime(df['date'].apply(lambda t: t.replace('\ufeff', '')), format="%Y-%m-%d")
# Set the time column as the index and drop redundant time column now that time is indexed. Do this op in-place.
df = df.set_index(df.date)
df.drop('date', inplace=True, axis=1)
# Group all events by (year, month) and count category by values.
counted_events = df.groupby([(df.index.year), (df.index.month)], as_index=True).category.value_counts()
counted_events.index.names = ["year", "month", "category"]
return counted_events
which gives me the following:
year month category
2017 1 A 2
B 1
2 C 3
A 1
The process to sum up all A's and B's would be quite manual since category becomes a part of the index in this case.
I'm an absolute pandas menace, so I'm likely making this much harder than it actually is. Can anyone give tips for how to achieve this grouping in pandas?
I tried this so posting though I like #Scott Boston's solution better as I combined A and B values earlier.
df.date = pd.to_datetime(df.date, format = '%Y-%m-%d')
df.loc[(df.category == 'A')|(df.category == 'B'), 'category'] = 'AB'
new_df = df.groupby([df.date.dt.year,df.date.dt.month]).category.value_counts().unstack().fillna(0)
new_df.columns = ['a_or_b_count', 'c_count']
new_df.index.names = ['Year', 'Month']
a_or_b_count c_count
Year Month
2017 1 3.0 0.0
2 1.0 3.0

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