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Python 3 match values based on column name similarity

I have a dataframe of the following form:
Year 1 Grade
Year 2 Grade
Year 3 Grade
Year 4 Grade
Year 1 Students
Year 2 Students
Year 3 Students
Year 4 Students
60
70
80
100
20
32
18
25
I would like to somehow transpose this table to the following format:
Year
Grade
Students
1
60
20
2
70
32
3
80
18
4
100
25
I created a list of years and initiated a new dataframe with the "year" column. I was thinking of matching the year integer to the column name containing it in the original DF, match and assign the correct value, but got stuck there.

You need a manual reshaping using a split of the Index into a MultiIndex:
out = (df
.set_axis(df.columns.str.split(expand=True), axis=1) # make MultiIndex
.iloc[0] # select row as Series
.unstack() # unstack Grade/Students
.droplevel(0) # remove literal "Year"
.rename_axis('Year') # set index name
.reset_index() # index to column
)
output:
Year Grade Students
0 1 60 20
1 2 70 32
2 3 80 18
3 4 100 25
Or using pivot_longer from janitor:
# pip install pyjanitor
import janitor
out = (df.pivot_longer(
names_to = ('ignore', 'Year', '.value'),
names_sep = ' ')
.drop(columns='ignore')
)
out
Year Grade Students
0 1 60 20
1 2 70 32
2 3 80 18
3 4 100 25
The .value determines which parts of the sub labels in the columns are retained; the labels are split apart by names_sep, which can be a string or a regex. Another option is by using a regex, with names_pattern to split and reshape the columns:
df.pivot_longer(names_to = ('Year', '.value'),
names_pattern = r'.+(\d)\s(.+)')
Year Grade Students
0 1 60 20
1 2 70 32
2 3 80 18
3 4 100 25

Here's one way to do it. Feel free to ask questions about how it works.
import pandas as pd
cols = ["Year 1 Grade", "Year 2 Grade", "Year 3 Grade" , "Year 4 Grade",
"Year 1 Students", "Year 2 Students", "Year 3 Students", "Year 4 Students"]
vals = [60,70,80,100,20,32,18,25]
vals = [[v] for v in vals]
df = pd.DataFrame({k:v for k,v in zip(cols,vals)})
grades = df.filter(like="Grade").T.reset_index(drop=True).rename(columns={0:"Grades"})
students = df.filter(like="Student").T.reset_index(drop=True).rename(columns={0:"Students"})
pd.concat([grades,students], axis=1)

I came up with these. grades here are your first row.
df = pd.DataFrame(grades) # your dataframe without columns
grades = np.array(df.iloc[0]).reshape(4,2) # extract the first row, turn it into an array and reshape it to 4*2
new_df = pd.DataFrame(grades).reset_index()
new_df.columns = ['Year', 'Grade', 'Students'] # rename the columns

Python Merging data frames

In python, I have a df that looks like this
Name ID
Anna 1
Polly 1
Sarah 2
Max 3
Kate 3
Ally 3
Steve 3
And a df that looks like this
Name ID
Dan 1
Hallie 2
Cam 2
Lacy 2
Ryan 3
Colt 4
Tia 4
How can I merge the df’s so that the ID column looks like this
Name ID
Anna 1
Polly 1
Sarah 2
Max 3
Kate 3
Ally 3
Steve 3
Dan 4
Hallie 5
Cam 5
Lacy 5
Ryan 6
Colt 7
Tia 7
This is just a minimal reproducible example. My actual data set has 1000’s of values. I’m basically merging data frames and want the ID’s in numerical order (continuation of previous data frame) instead of repeating from one each time. I know that I can reset the index if ID is a unique identifier. But in this case, more than one person can have the same ID. So how can I account for that?

From the example that you have provided above, you can observe that we can obtain the final dataframe by: adding the maximum value of ID in first df to the second and then concatenating them, to explain this better:
Name df2 final_df
Dan 1 4
This value in final_df is obtained by doing a 1+(max value of ID from df1 i.e. 3) and this trend is followed for all entries for the dataframe.
Code:
import pandas as pd
df = pd.DataFrame({'Name':['Anna','Polly','Sarah','Max','Kate','Ally','Steve'],'ID':[1,1,2,3,3,3,3]})
df1 = pd.DataFrame({'Name':['Dan','Hallie','Cam','Lacy','Ryan','Colt','Tia'],'ID':[1,2,2,2,3,4,4]})
max_df = df['ID'].max()
df1['ID'] = df1['ID'].apply(lambda x: x+max_df)
final_df = pd.concat([df,df1])
print(final_df)

How to findout difference between two dataframes irrespective of index? [duplicate]

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.

Check if row value in a dataframe exists in another dataframe using loop for reconciliation

I am looking to develop some generic logic that will allow me to perform reconciliation between 2 datasets.
I have 2 dataframes and I want to loop through every row value in df1 and check if it exists in df2. If it does exist I want to create a new column 'Match' in df1 with the value 'Yes' and if it does not exist I want to append the missing values in a separate df which I will print to csv.
Example datasets:
df1:
ID Name Age
1 Adam 45
2 Bill 44
3 Claire 23
df2:
ID Name Age
1 Adam 45
2 Bill 44
3 Claire 23
4 Bob 40
5 Chris 21
The column names in the 2 dataframes I've used here are just for reference. But essentially I want to check if the row (1, Adam, 45) in df1 exists in df2.
The output for df3 would look like this:
df3:
ID Name Age
4 Bob 40
5 Chris 21
The updated df1 would look like this:
df2:
ID Name Age Match
1 Adam 45 Yes
2 Bill 44 Yes
3 Claire 23 Yes
To be clear, I understand that this can be done using a merge or isin, but would like a fluid solution that can be used for any dataset.
I appreciate this might be a big ask as I haven't provided much guidline but any help with this would be great!!
Thanks!!

You need to use merge here and utilize the indicator=True feature:
df_all = df1.merge(df2, on=['ID'], how='outer', indicator=True)
df3 = df_all[df_all['_merge'] == 'right_only'].drop(columns=['Name_x', 'Age_x']).rename(columns={'Name_y': 'Name', 'Age_y': 'Age'})[['ID', 'Name', 'Age']]
df2 = df_all[df_all['_merge'] == 'both'].drop(columns=['Name_x', 'Age_x']).rename(columns={'Name_y': 'Name', 'Age_y': 'Age'})[['ID', 'Name', 'Age']]
print(df3)
print(df2)
df3:
ID Name Age
3 4 Bob 40
4 5 Chris 21
df2:
ID Name Age
0 1 Adam 45
1 2 Bill 44
2 3 Claire 23

Merging rows from different dataframes together

I have two Dataframes: one with columns "Name", "Year" and "Type" and the other one with different parameters. There are 4 different types and each one has his specific parameters. Now i need to merge them together.
My approach is to use a if-function to find out the "type". For example in row two of df3 i have type 'a'. The parameters for type 'a' are in row 3 of df4. I tried to connect them with the following code:
df3.ix[[2]]
s1 = df3.ix[[2]]
s2 = df4.ix[[3]]
result = pd.concat([s1, s2], axis=1)
My problem is now, that the parameters are in a seperate row and not added to row 2. Is there a chance to merge them together in one row? Thanks for your answers!

If df3 has a Type column and df4 has a type column, then the two DataFrames can be merged with
pd.merge(df3, df4, left_on='Type', right_on='type')
This is by default an inner join.
In [13]: df3
Out[13]:
Name Year Type
1 A 2012 boat
2 B 2013 car
3 C 2011 truck
4 D 2013 boat
In [14]: df4
Out[14]:
type Parameter1 Parameter2 Parameter3
0 boat 2 8 7
1 car 1 9 3
2 truck 5 4 2
In [15]: pd.merge(df3, df4, left_on='Type', right_on='type')
Out[15]:
Name Year Type type Parameter1 Parameter2 Parameter3
0 A 2012 boat boat 2 8 7
1 D 2013 boat boat 2 8 7
2 B 2013 car car 1 9 3
3 C 2011 truck truck 5 4 2
Note that if the column names matched exactly, then
pd.merge(df3, df4)
would merge on column names shared in common by default.