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 the following data frames:
print(df_a)
mukey DI PI
0 100000 35 14
1 1000005 44 14
2 1000006 44 14
3 1000007 43 13
4 1000008 43 13
print(df_b)
mukey niccdcd
0 190236 4
1 190237 6
2 190238 7
3 190239 4
4 190240 7
When I try to join these data frames:
join_df = df_a.join(df_b, on='mukey', how='left')
I get the error:
*** ValueError: columns overlap but no suffix specified: Index([u'mukey'], dtype='object')
Why is this so? The data frames do have common 'mukey' values.
Your error on the snippet of data you posted is a little cryptic, in that because there are no common values, the join operation fails because the values don't overlap it requires you to supply a suffix for the left and right hand side:
In [173]:
df_a.join(df_b, on='mukey', how='left', lsuffix='_left', rsuffix='_right')
Out[173]:
mukey_left DI PI mukey_right niccdcd
index
0 100000 35 14 NaN NaN
1 1000005 44 14 NaN NaN
2 1000006 44 14 NaN NaN
3 1000007 43 13 NaN NaN
4 1000008 43 13 NaN NaN
merge works because it doesn't have this restriction:
In [176]:
df_a.merge(df_b, on='mukey', how='left')
Out[176]:
mukey DI PI niccdcd
0 100000 35 14 NaN
1 1000005 44 14 NaN
2 1000006 44 14 NaN
3 1000007 43 13 NaN
4 1000008 43 13 NaN
The .join() function is using the index of the passed as argument dataset, so you should use set_index or use .merge function instead.
Please find the two examples that should work in your case:
join_df = LS_sgo.join(MSU_pi.set_index('mukey'), on='mukey', how='left')
or
join_df = df_a.merge(df_b, on='mukey', how='left')
This error indicates that the two tables have one or more column names that have the same column name.
The error message translates to: "I can see the same column in both tables but you haven't told me to rename either one before bringing them into the same table"
You either want to delete one of the columns before bringing it in from the other on using del df['column name'], or use lsuffix to re-write the original column, or rsuffix to rename the one that is being brought in.
df_a.join(df_b, on='mukey', how='left', lsuffix='_left', rsuffix='_right')
The error indicates that the two tables have the 1 or more column names that have the same column name.
Anyone with the same error who doesn't want to provide a suffix can rename the columns instead. Also make sure the index of both DataFrames match in type and value if you don't want to provide the on='mukey' setting.
# rename example
df_a = df_a.rename(columns={'a_old': 'a_new', 'a2_old': 'a2_new'})
# set the index
df_a = df_a.set_index(['mukus'])
df_b = df_b.set_index(['mukus'])
df_a.join(df_b)
Mainly join is used exclusively to join based on the index,not on the attribute names,so change the attributes names in two different dataframes,then try to join,they will be joined,else this error is raised
This question already has answers here:
How to change the order of DataFrame columns?
(41 answers)
Closed 3 years ago.
I have a dataframe with over 200 columns. The issue is as they were generated the order is
['Q1.3','Q6.1','Q1.2','Q1.1',......]
I need to sort the columns as follows:
['Q1.1','Q1.2','Q1.3',.....'Q6.1',......]
Is there some way for me to do this within Python?
df = df.reindex(sorted(df.columns), axis=1)
This assumes that sorting the column names will give the order you want. If your column names won't sort lexicographically (e.g., if you want column Q10.3 to appear after Q9.1), you'll need to sort differently, but that has nothing to do with pandas.
You can also do more succinctly:
df.sort_index(axis=1)
Make sure you assign the result back:
df = df.sort_index(axis=1)
Or, do it in-place:
df.sort_index(axis=1, inplace=True)
You can just do:
df[sorted(df.columns)]
Edit: Shorter is
df[sorted(df)]
For several columns, You can put columns order what you want:
#['A', 'B', 'C'] <-this is your columns order
df = df[['C', 'B', 'A']]
This example shows sorting and slicing columns:
d = {'col1':[1, 2, 3], 'col2':[4, 5, 6], 'col3':[7, 8, 9], 'col4':[17, 18, 19]}
df = pandas.DataFrame(d)
You get:
col1 col2 col3 col4
1 4 7 17
2 5 8 18
3 6 9 19
Then do:
df = df[['col3', 'col2', 'col1']]
Resulting in:
col3 col2 col1
7 4 1
8 5 2
9 6 3
Tweet's answer can be passed to BrenBarn's answer above with
data.reindex_axis(sorted(data.columns, key=lambda x: float(x[1:])), axis=1)
So for your example, say:
vals = randint(low=16, high=80, size=25).reshape(5,5)
cols = ['Q1.3', 'Q6.1', 'Q1.2', 'Q9.1', 'Q10.2']
data = DataFrame(vals, columns = cols)
You get:
data
Q1.3 Q6.1 Q1.2 Q9.1 Q10.2
0 73 29 63 51 72
1 61 29 32 68 57
2 36 49 76 18 37
3 63 61 51 30 31
4 36 66 71 24 77
Then do:
data.reindex_axis(sorted(data.columns, key=lambda x: float(x[1:])), axis=1)
resulting in:
data
Q1.2 Q1.3 Q6.1 Q9.1 Q10.2
0 2 0 1 3 4
1 7 5 6 8 9
2 2 0 1 3 4
3 2 0 1 3 4
4 2 0 1 3 4
If you need an arbitrary sequence instead of sorted sequence, you could do:
sequence = ['Q1.1','Q1.2','Q1.3',.....'Q6.1',......]
your_dataframe = your_dataframe.reindex(columns=sequence)
I tested this in 2.7.10 and it worked for me.
Don't forget to add "inplace=True" to Wes' answer or set the result to a new DataFrame.
df.sort_index(axis=1, inplace=True)
The quickest method is:
df.sort_index(axis=1)
Be aware that this creates a new instance. Therefore you need to store the result in a new variable:
sortedDf=df.sort_index(axis=1)
The sort method and sorted function allow you to provide a custom function to extract the key used for comparison:
>>> ls = ['Q1.3', 'Q6.1', 'Q1.2']
>>> sorted(ls, key=lambda x: float(x[1:]))
['Q1.2', 'Q1.3', 'Q6.1']
One use-case is that you have named (some of) your columns with some prefix, and you want the columns sorted with those prefixes all together and in some particular order (not alphabetical).
For example, you might start all of your features with Ft_, labels with Lbl_, etc, and you want all unprefixed columns first, then all features, then the label. You can do this with the following function (I will note a possible efficiency problem using sum to reduce lists, but this isn't an issue unless you have a LOT of columns, which I do not):
def sortedcols(df, groups = ['Ft_', 'Lbl_'] ):
return df[ sum([list(filter(re.compile(r).search, list(df.columns).copy())) for r in (lambda l: ['^(?!(%s))' % '|'.join(l)] + ['^%s' % i for i in l ] )(groups) ], []) ]
print df.sort_index(by='Frequency',ascending=False)
where by is the name of the column,if you want to sort the dataset based on column