I am using the code below to make a search on a .csv file and match a column in both files and grab a different column I want and add it as a new column. However, I am trying to make the match based on two columns instead of one. Is there a way to do this?
import pandas as pd
df1 = pd.read_csv("matchone.csv")
df2 = pd.read_csv("comingfrom.csv")
def lookup_prod(ip):
for row in df2.itertuples():
if ip in row[1]:
return row[3]
else:
return '0'
df1['want'] = df1['name'].apply(lookup_prod)
df1[df1.want != '0']
print(df1)
#df1.to_csv('file_name.csv')
The code above makes a search from the column name 'samename' in both files and gets the column I request ([3]) from the df2. I want to make the code make a match for both column 'name' and another column 'price' and only if both columns in both df1 and df2 match then the code take the value on ([3]).
df 1 :
name price value
a 10 35
b 10 21
c 10 33
d 10 20
e 10 88
df 2 :
name price want
a 10 123
b 5 222
c 10 944
d 10 104
e 5 213
When the code is run (asking for the want column from d2, based on both if df1 name = df2 name) the produced result is :
name price value want
a 10 35 123
b 10 21 222
c 10 33 944
d 10 20 104
e 10 88 213
However, what I want is if both df1 name = df2 name and df1 price = df2 price, then take the column df2 want, so the desired result is:
name price value want
a 10 35 123
b 10 21 0
c 10 33 944
d 10 20 104
e 10 88 0
You need to use pandas.DataFrame.merge() method with multiple keys:
df1.merge(df2, on=['name','price'], how='left').fillna(0)
Method represents missing values as NaNs, so that the column's dtype changes to float64 but you can change it back after filling the missed values with 0.
Also please be aware that duplicated combinations of name and price in df2 will appear several times in the result.
If you are matching the two dataframes based on the name and the price, you can use df.where and df.isin
df1['want'] = df2['want'].where(df1[['name','price']].isin(df2).all(axis=1)).fillna('0')
df1
name price value want
0 a 10 35 123.0
1 b 10 21 0
2 c 10 33 944.0
3 d 10 20 104.0
4 e 10 88 0
Expanding on https://stackoverflow.com/a/73830294/20110802:
You can add the validate option to the merge in order to avoid duplication on one side (or both):
pd.merge(df1, df2, on=['name','price'], how='left', validate='1:1').fillna(0)
Also, if the float conversion is a problem for you, one option is to do an inner join first and then pd.concat the result with the "leftover" df1 where you already added a constant valued column. Would look something like:
df_inner = pd.merge(df1, df2, on=['name', 'price'], how='inner', validate='1:1')
merged_pairs = set(zip(df_inner.name, df_inner.price))
df_anti = df1.loc[~pd.Series(zip(df1.name, df1.price)).isin(merged_pairs)]
df_anti['want'] = 0
df_result = pd.concat([df_inner, df_anti]) # perhaps ignore_index=True ?
Looks complicated, but should be quite performant because it filters by set. I think there might be a possibility to set name and price as index, merge on index and then filter by index to not having to do the zip-set-shenanigans, bit I'm no expert on multiindex-handling.
#Try this code it will give you expected results
import pandas as pd
df1 = pd.DataFrame({'name' :['a','b','c','d','e'] ,
'price' :[10,10,10,10,10],
'value' : [35,21,33,20,88]})
df2 = pd.DataFrame({'name' :['a','b','c','d','e'] ,
'price' :[10,5,10,10,5],
'want' : [123,222,944,104 ,213]})
new = pd.merge(df1,df2, how='left', left_on=['name','price'], right_on=['name','price'])
print(new.fillna(0))
I have two dataframes, df1 and df2. df1 has repeat observations arranged in wide format, and df2 in long format.
import pandas as pd
df1 = pd.DataFrame({"ID":[1,2,3],"colA_1":[1,2,3],"date1":["1.1.2001", "2.1.2001","3.1.2001"],"colA_2":[4,5,6],"date2":["1.1.2002", "2.1.2002","3.1.2002"]})
df2 = pd.DataFrame({"ID":[1,1,2,2,3,3],"col1":[1,1.5,2,2.5,3,3.5],"date":["1.1.2001", "1.1.2002","2.1.2001","2.1.2002","3.1.2001","3.1.2002"], "col3":[11,12,13,14,15,16],"col4":[21,22,23,24,25,26]})
df1 looks like:
ID colA_1 date1 colA_2 date2
0 1 1 1.1.2001 4 1.1.2002
1 2 2 2.1.2001 5 2.1.2002
2 3 3 3.1.2001 6 3.1.2002
df2 looks like:
ID col1 date1 col3 col4
0 1 1.0 1.1.2001 11 21
1 1 1.5 1.1.2002 12 22
2 2 2.0 2.1.2001 13 23
3 2 2.5 2.1.2002 14 24
4 3 3.0 3.1.2001 15 25
5 3 3.5 3.1.2002 16 26
6 3 4.0 4.1.2002 17 27
I want to take a given column from df2, "col3", and then:
(1) if the columns "ID" and "date" in df2 match with the columns "ID" and "date1" in df1, I want to put the value in a new column in df1 called "colB_1".
(2) else if the columns "ID" and "date" in df2 match with the columns "ID" and "date2" in df1, I want to put the value in a new column in df1 called "colB_2".
(3) else if the columns "ID" and "date" in df2 have no match with either ("ID" and "date1") or ("ID" and "date2"), I want to ignore these rows.
So, the output of this output dataframe, df3, should look like this:
ID colA_1 date1 colA_2 date2 colB_1 colB_2
0 1 1 1.1.2001 4 1.1.2002 11 12
1 2 2 2.1.2001 5 2.1.2002 13 14
2 3 3 3.1.2001 6 3.1.2002 15 16
What is the best way to do this?
I found this link, but the answer doesn't work for my case. I would like a really explicit way to specify column matching. I think it's possible that df.mask might be able to help me, but I am not sure how to implement it.
e.g.: the following code
df3 = df1.copy()
df3["colB_1"] = ""
df3["colB_2"] = ""
filter1 = (df1["ID"] == df2["ID"]) & (df1["date1"] == df2["date"])
filter2 = (df1["ID"] == df2["ID"]) & (df1["date2"] == df2["date"])
df3["colB_1"] = df.mask(filter1, other=df2["col3"])
df3["colB_2"] = df.mask(filter2, other=df2["col3"])
gives the error
ValueError: Can only compare identically-labeled Series objects
I asked this question previously, and it was marked as closed; my question was marked as a duplicate of this one. However, this is not the case. The answers in the linked question suggest the use of either map or df.merge. Map does not work with multiple conditions (in my case, ID and date). And df.merge (the answer given for matching multiple columns) does not work in my case when one of the column names in df1 and df2 that are to be merged are different ("date" and "date1", for example).
For example, the below code:
df3 = df1.merge(df2[["ID","date","col3"]], on=['ID','date1'], how='left')
fails with a Key Error.
Also noteworthy is that I will be dealing with many different files, with many different column naming schemes, and I will need a different subset each time. This is why I would like an answer that explicitly names the columns and conditions.
Any help with this would be much appreciated.
You can the pd.wide_to_long after replacing the underscore , this will unpivot the dataframe which you can use to merge with df2 and then pivot back using unstack:
m =df1.rename(columns=lambda x: x.replace('_',''))
unpiv = pd.wide_to_long(m,['colA','date'],'ID','v').reset_index()
merge_piv = (unpiv.merge(df2[['ID','date','col3']],on=['ID','date'],how='left')
.set_index(['ID','v'])['col3'].unstack().add_prefix('colB_'))
final = df1.merge(merge_piv,left_on='ID',right_index=True)
ID colA_1 date1 colA_2 date2 colB_1 colB_2
0 1 1 1.1.2001 4 1.1.2002 11 12
1 2 2 2.1.2001 5 2.1.2002 13 14
2 3 3 3.1.2001 6 3.1.2002 15 16
I have two data frames df1 and df2, where df2 is a subset of df1. How do I get a new data frame (df3) which is the difference between the two data frames?
In other word, a data frame that has all the rows/columns in df1 that are not in df2?
By using drop_duplicates
pd.concat([df1,df2]).drop_duplicates(keep=False)
Update :
The above method only works for those data frames that don't already have duplicates themselves. For example:
df1=pd.DataFrame({'A':[1,2,3,3],'B':[2,3,4,4]})
df2=pd.DataFrame({'A':[1],'B':[2]})
It will output like below , which is wrong
Wrong Output :
pd.concat([df1, df2]).drop_duplicates(keep=False)
Out[655]:
A B
1 2 3
Correct Output
Out[656]:
A B
1 2 3
2 3 4
3 3 4
How to achieve that?
Method 1: Using isin with tuple
df1[~df1.apply(tuple,1).isin(df2.apply(tuple,1))]
Out[657]:
A B
1 2 3
2 3 4
3 3 4
Method 2: merge with indicator
df1.merge(df2,indicator = True, how='left').loc[lambda x : x['_merge']!='both']
Out[421]:
A B _merge
1 2 3 left_only
2 3 4 left_only
3 3 4 left_only
For rows, try this, where Name is the joint index column (can be a list for multiple common columns, or specify left_on and right_on):
m = df1.merge(df2, on='Name', how='outer', suffixes=['', '_'], indicator=True)
The indicator=True setting is useful as it adds a column called _merge, with all changes between df1 and df2, categorized into 3 possible kinds: "left_only", "right_only" or "both".
For columns, try this:
set(df1.columns).symmetric_difference(df2.columns)
Accepted answer Method 1 will not work for data frames with NaNs inside, as pd.np.nan != pd.np.nan. I am not sure if this is the best way, but it can be avoided by
df1[~df1.astype(str).apply(tuple, 1).isin(df2.astype(str).apply(tuple, 1))]
It's slower, because it needs to cast data to string, but thanks to this casting pd.np.nan == pd.np.nan.
Let's go trough the code. First we cast values to string, and apply tuple function to each row.
df1.astype(str).apply(tuple, 1)
df2.astype(str).apply(tuple, 1)
Thanks to that, we get pd.Series object with list of tuples. Each tuple contains whole row from df1/df2.
Then we apply isin method on df1 to check if each tuple "is in" df2.
The result is pd.Series with bool values. True if tuple from df1 is in df2. In the end, we negate results with ~ sign, and applying filter on df1. Long story short, we get only those rows from df1 that are not in df2.
To make it more readable, we may write it as:
df1_str_tuples = df1.astype(str).apply(tuple, 1)
df2_str_tuples = df2.astype(str).apply(tuple, 1)
df1_values_in_df2_filter = df1_str_tuples.isin(df2_str_tuples)
df1_values_not_in_df2 = df1[~df1_values_in_df2_filter]
import pandas as pd
# given
df1 = pd.DataFrame({'Name':['John','Mike','Smith','Wale','Marry','Tom','Menda','Bolt','Yuswa',],
'Age':[23,45,12,34,27,44,28,39,40]})
df2 = pd.DataFrame({'Name':['John','Smith','Wale','Tom','Menda','Yuswa',],
'Age':[23,12,34,44,28,40]})
# find elements in df1 that are not in df2
df_1notin2 = df1[~(df1['Name'].isin(df2['Name']) & df1['Age'].isin(df2['Age']))].reset_index(drop=True)
# output:
print('df1\n', df1)
print('df2\n', df2)
print('df_1notin2\n', df_1notin2)
# df1
# Age Name
# 0 23 John
# 1 45 Mike
# 2 12 Smith
# 3 34 Wale
# 4 27 Marry
# 5 44 Tom
# 6 28 Menda
# 7 39 Bolt
# 8 40 Yuswa
# df2
# Age Name
# 0 23 John
# 1 12 Smith
# 2 34 Wale
# 3 44 Tom
# 4 28 Menda
# 5 40 Yuswa
# df_1notin2
# Age Name
# 0 45 Mike
# 1 27 Marry
# 2 39 Bolt
Perhaps a simpler one-liner, with identical or different column names. Worked even when df2['Name2'] contained duplicate values.
newDf = df1.set_index('Name1')
.drop(df2['Name2'], errors='ignore')
.reset_index(drop=False)
edit2, I figured out a new solution without the need of setting index
newdf=pd.concat([df1,df2]).drop_duplicates(keep=False)
Okay i found the answer of highest vote already contain what I have figured out. Yes, we can only use this code on condition that there are no duplicates in each two dfs.
I have a tricky method. First we set ’Name’ as the index of two dataframe given by the question. Since we have same ’Name’ in two dfs, we can just drop the ’smaller’ df’s index from the ‘bigger’ df.
Here is the code.
df1.set_index('Name',inplace=True)
df2.set_index('Name',inplace=True)
newdf=df1.drop(df2.index)
Pandas now offers a new API to do data frame diff: pandas.DataFrame.compare
df.compare(df2)
col1 col3
self other self other
0 a c NaN NaN
2 NaN NaN 3.0 4.0
In addition to accepted answer, I would like to propose one more wider solution that can find a 2D set difference of two dataframes with any index/columns (they might not coincide for both datarames). Also method allows to setup tolerance for float elements for dataframe comparison (it uses np.isclose)
import numpy as np
import pandas as pd
def get_dataframe_setdiff2d(df_new: pd.DataFrame,
df_old: pd.DataFrame,
rtol=1e-03, atol=1e-05) -> pd.DataFrame:
"""Returns set difference of two pandas DataFrames"""
union_index = np.union1d(df_new.index, df_old.index)
union_columns = np.union1d(df_new.columns, df_old.columns)
new = df_new.reindex(index=union_index, columns=union_columns)
old = df_old.reindex(index=union_index, columns=union_columns)
mask_diff = ~np.isclose(new, old, rtol, atol)
df_bool = pd.DataFrame(mask_diff, union_index, union_columns)
df_diff = pd.concat([new[df_bool].stack(),
old[df_bool].stack()], axis=1)
df_diff.columns = ["New", "Old"]
return df_diff
Example:
In [1]
df1 = pd.DataFrame({'A':[2,1,2],'C':[2,1,2]})
df2 = pd.DataFrame({'A':[1,1],'B':[1,1]})
print("df1:\n", df1, "\n")
print("df2:\n", df2, "\n")
diff = get_dataframe_setdiff2d(df1, df2)
print("diff:\n", diff, "\n")
Out [1]
df1:
A C
0 2 2
1 1 1
2 2 2
df2:
A B
0 1 1
1 1 1
diff:
New Old
0 A 2.0 1.0
B NaN 1.0
C 2.0 NaN
1 B NaN 1.0
C 1.0 NaN
2 A 2.0 NaN
C 2.0 NaN
As mentioned here
that
df1[~df1.apply(tuple,1).isin(df2.apply(tuple,1))]
is correct solution but it will produce wrong output if
df1=pd.DataFrame({'A':[1],'B':[2]})
df2=pd.DataFrame({'A':[1,2,3,3],'B':[2,3,4,4]})
In that case above solution will give
Empty DataFrame, instead you should use concat method after removing duplicates from each datframe.
Use concate with drop_duplicates
df1=df1.drop_duplicates(keep="first")
df2=df2.drop_duplicates(keep="first")
pd.concat([df1,df2]).drop_duplicates(keep=False)
I had issues with handling duplicates when there were duplicates on one side and at least one on the other side, so I used Counter.collections to do a better diff, ensuring both sides have the same count. This doesn't return duplicates, but it won't return any if both sides have the same count.
from collections import Counter
def diff(df1, df2, on=None):
"""
:param on: same as pandas.df.merge(on) (a list of columns)
"""
on = on if on else df1.columns
df1on = df1[on]
df2on = df2[on]
c1 = Counter(df1on.apply(tuple, 'columns'))
c2 = Counter(df2on.apply(tuple, 'columns'))
c1c2 = c1-c2
c2c1 = c2-c1
df1ondf2on = pd.DataFrame(list(c1c2.elements()), columns=on)
df2ondf1on = pd.DataFrame(list(c2c1.elements()), columns=on)
df1df2 = df1.merge(df1ondf2on).drop_duplicates(subset=on)
df2df1 = df2.merge(df2ondf1on).drop_duplicates(subset=on)
return pd.concat([df1df2, df2df1])
> df1 = pd.DataFrame({'a': [1, 1, 3, 4, 4]})
> df2 = pd.DataFrame({'a': [1, 2, 3, 4, 4]})
> diff(df1, df2)
a
0 1
0 2
There is a new method in pandas DataFrame.compare that compare 2 different dataframes and return which values changed in each column for the data records.
Example
First Dataframe
Id Customer Status Date
1 ABC Good Mar 2023
2 BAC Good Feb 2024
3 CBA Bad Apr 2022
Second Dataframe
Id Customer Status Date
1 ABC Bad Mar 2023
2 BAC Good Feb 2024
5 CBA Good Apr 2024
Comparing Dataframes
print("Dataframe difference -- \n")
print(df1.compare(df2))
print("Dataframe difference keeping equal values -- \n")
print(df1.compare(df2, keep_equal=True))
print("Dataframe difference keeping same shape -- \n")
print(df1.compare(df2, keep_shape=True))
print("Dataframe difference keeping same shape and equal values -- \n")
print(df1.compare(df2, keep_shape=True, keep_equal=True))
Result
Dataframe difference --
Id Status Date
self other self other self other
0 NaN NaN Good Bad NaN NaN
2 3.0 5.0 Bad Good Apr 2022 Apr 2024
Dataframe difference keeping equal values --
Id Status Date
self other self other self other
0 1 1 Good Bad Mar 2023 Mar 2023
2 3 5 Bad Good Apr 2022 Apr 2024
Dataframe difference keeping same shape --
Id Customer Status Date
self other self other self other self other
0 NaN NaN NaN NaN Good Bad NaN NaN
1 NaN NaN NaN NaN NaN NaN NaN NaN
2 3.0 5.0 NaN NaN Bad Good Apr 2022 Apr 2024
Dataframe difference keeping same shape and equal values --
Id Customer Status Date
self other self other self other self other
0 1 1 ABC ABC Good Bad Mar 2023 Mar 2023
1 2 2 BAC BAC Good Good Feb 2024 Feb 2024
2 3 5 CBA CBA Bad Good Apr 2022 Apr 2024
A slight variation of the nice #liangli's solution that does not require to change the index of existing dataframes:
newdf = df1.drop(df1.join(df2.set_index('Name').index))
Finding difference by index. Assuming df1 is a subset of df2 and the indexes are carried forward when subsetting
df1.loc[set(df1.index).symmetric_difference(set(df2.index))].dropna()
# Example
df1 = pd.DataFrame({"gender":np.random.choice(['m','f'],size=5), "subject":np.random.choice(["bio","phy","chem"],size=5)}, index = [1,2,3,4,5])
df2 = df1.loc[[1,3,5]]
df1
gender subject
1 f bio
2 m chem
3 f phy
4 m bio
5 f bio
df2
gender subject
1 f bio
3 f phy
5 f bio
df3 = df1.loc[set(df1.index).symmetric_difference(set(df2.index))].dropna()
df3
gender subject
2 m chem
4 m bio
Defining our dataframes:
df1 = pd.DataFrame({
'Name':
['John','Mike','Smith','Wale','Marry','Tom','Menda','Bolt','Yuswa'],
'Age':
[23,45,12,34,27,44,28,39,40]
})
df2 = df1[df1.Name.isin(['John','Smith','Wale','Tom','Menda','Yuswa'])
df1
Name Age
0 John 23
1 Mike 45
2 Smith 12
3 Wale 34
4 Marry 27
5 Tom 44
6 Menda 28
7 Bolt 39
8 Yuswa 40
df2
Name Age
0 John 23
2 Smith 12
3 Wale 34
5 Tom 44
6 Menda 28
8 Yuswa 40
The difference between the two would be:
df1[~df1.isin(df2)].dropna()
Name Age
1 Mike 45.0
4 Marry 27.0
7 Bolt 39.0
Where:
df1.isin(df2) returns the rows in df1 that are also in df2.
~ (Element-wise logical NOT) in front of the expression negates the results, so we get the elements in df1 that are NOT in df2–the difference between the two.
.dropna() drops the rows with NaN presenting the desired output
Note This only works if len(df1) >= len(df2). If df2 is longer than df1 you can reverse the expression: df2[~df2.isin(df1)].dropna()
I found the deepdiff library is a wonderful tool that also extends well to dataframes if different detail is required or ordering matters. You can experiment with diffing to_dict('records'), to_numpy(), and other exports:
import pandas as pd
from deepdiff import DeepDiff
df1 = pd.DataFrame({
'Name':
['John','Mike','Smith','Wale','Marry','Tom','Menda','Bolt','Yuswa'],
'Age':
[23,45,12,34,27,44,28,39,40]
})
df2 = df1[df1.Name.isin(['John','Smith','Wale','Tom','Menda','Yuswa'])]
DeepDiff(df1.to_dict(), df2.to_dict())
# {'dictionary_item_removed': [root['Name'][1], root['Name'][4], root['Name'][7], root['Age'][1], root['Age'][4], root['Age'][7]]}
Symmetric Difference
If you are interested in the rows that are only in one of the dataframes but not both, you are looking for the set difference:
pd.concat([df1,df2]).drop_duplicates(keep=False)
⚠️ Only works, if both dataframes do not contain any duplicates.
Set Difference / Relational Algebra Difference
If you are interested in the relational algebra difference / set difference, i.e. df1-df2 or df1\df2:
pd.concat([df1,df2,df2]).drop_duplicates(keep=False)
⚠️ Only works, if both dataframes do not contain any duplicates.
Another possible solution is to use numpy broadcasting:
df1[np.all(~np.all(df1.values == df2.values[:, None], axis=2), axis=0)]
Output:
Name Age
1 Mike 45
4 Marry 27
7 Bolt 39
Using the lambda function you can filter the rows with _merge value “left_only” to get all the rows in df1 which are missing from df2
df3 = df1.merge(df2, how = 'outer' ,indicator=True).loc[lambda x :x['_merge']=='left_only']
df
Try this one:
df_new = df1.merge(df2, how='outer', indicator=True).query('_merge == "left_only"').drop('_merge', 1)
It will result a new dataframe with the differences: the values that exist in df1 but not in df2.
I have 2 dataframe and I need to get new column to first dataframe, using values from second
FIrse df is
ID,"url","used_at","active_seconds"
8075643aab791cec7dc9d18926958b67,"sberbank.ru/ru/person/promo/10mnl?utm_source=Vesti.ru&utm_medium=html&utm_campaign=10_million_users_SBOL_dec2015&utm_term=every14_syncbanners",2016-01-01 00:03:16,183
a04a8041ffa6fe1b85471ca5af1ee575,"online.rsb.ru/hb/faces/system/login/rslogin.jsp?credit=false",2016-01-01 00:04:36,42
a04a8041ffa6fe1b85471ca5af1ee575,"online.rsb.ru/hb/faces/system/login/sms/sms.jsp?smsAuth=true",2016-01-01 00:05:18,22
a04a8041ffa6fe1b85471ca5af1ee575,"online.rsb.ru/hb/faces/rs/RSIndex.jspx",2016-01-01 00:05:40,14
a04a8041ffa6fe1b85471ca5af1ee575,"online.rsb.ru/hb/faces/rs/payments/PaymentReq.jspx",2016-01-01 00:05:54,22
ba880911a6d54f6ea6d3145081a0e0dd,"homecredit.ru/help/quest/feedback.php",2016-01-01 00:06:12,2
Second df looks like
URL Code
citibank\.ru\/russia\/info\/rus\/contacts_form\.htm 15
citibank\.ru\/russia\/info\/rus\/contacts\.htm 15
gazprombank\.ru\/contacts\/ 15
gazprombank\.ru\/feedback\/ 15
gazprombank\.ru\/additional_office\/ 15
homecredit\.ru\/help\/quest\/feedback\.php 15
homecredit\.ru\/offices\/* 15
If I don't have a regex, I use
df1['code'] = df1.url.map(df2.set_index('URL')['Code'])
But I can't do this, because df2.URL is regex.
But
df1['code'] = df1['url'].replace(df2['URL'], df2['Code'], regex=True)
doesn't work.
As per my comment, the pandas.Series.replace() method doesn't allow Series objects as the to_replace and value arguments. Passing lists instead works:
df1['code'] = df1.url.replace(df2.URL.values, df2.Code.values, regex=True)
print df1[['url', 'code']]
produces the following output:
url \
0 sberbank.ru/ru/person/promo/10mnl?utm_source=V...
1 online.rsb.ru/hb/faces/system/login/rslogin.js...
2 online.rsb.ru/hb/faces/system/login/sms/sms.js...
3 online.rsb.ru/hb/faces/rs/RSIndex.jspx
4 online.rsb.ru/hb/faces/rs/payments/PaymentReq....
5 homecredit.ru/help/quest/feedback.php
code
0 sberbank.ru/ru/person/promo/10mnl?utm_source=V...
1 online.rsb.ru/hb/faces/system/login/rslogin.js...
2 online.rsb.ru/hb/faces/system/login/sms/sms.js...
3 online.rsb.ru/hb/faces/rs/RSIndex.jspx
4 online.rsb.ru/hb/faces/rs/payments/PaymentReq....
5 15
In answer to your additional comments, you can't get df2.Code in df1.code in rows where df1.url doesn't match any of the regex strings, but you can provide a value (e.g. None) for these cases to be put in the column instead. This is, for example, done by adding the following line:
df1['code'] = df1.apply(lambda x: None if x.code == x.url else x.code, axis=1)
where print df1[['url', 'code']] returns the following:
url code
0 sberbank.ru/ru/person/promo/10mnl?utm_source=V... NaN
1 online.rsb.ru/hb/faces/system/login/rslogin.js... NaN
2 online.rsb.ru/hb/faces/system/login/sms/sms.js... NaN
3 online.rsb.ru/hb/faces/rs/RSIndex.jspx NaN
4 online.rsb.ru/hb/faces/rs/payments/PaymentReq.... NaN
5 homecredit.ru/help/quest/feedback.php 15.0