Comparing two dataframes and getting the differences [duplicate]

问题描述：

I have two dataframes. Example:

df1:
Date       Fruit  Num  Color 
2013-11-24 Banana 22.1 Yellow
2013-11-24 Orange  8.6 Orange
2013-11-24 Apple   7.6 Green
2013-11-24 Celery 10.2 Green

df2:
Date       Fruit  Num  Color 
2013-11-24 Banana 22.1 Yellow
2013-11-24 Orange  8.6 Orange
2013-11-24 Apple   7.6 Green
2013-11-24 Celery 10.2 Green
2013-11-25 Apple  22.1 Red
2013-11-25 Orange  8.6 Orange

Each dataframe has the Date as an index. Both dataframes have the same structure.

What i want to do, is compare these two dataframes and find which rows are in df2 that aren't in df1. I want to compare the date (index) and the first column (Banana, APple, etc) to see if they exist in df2 vs df1.

I have tried the following:

• Compare two DataFrames and output their differences side-by-side

• Comparing two pandas dataframes for differences

For the first approach I get this error: "Exception: Can only compare identically-labeled DataFrame objects". I have tried removing the Date as index but get the same error.

On the third approach, I get the assert to return False but cannot figure out how to actually see the different rows.

Any pointers would be welcome

解决方案 1：

This approach, df1 != df2, works only for dataframes with identical rows and columns. In fact, all dataframes axes are compared with _indexed_same method, and exception is raised if differences found, even in columns/indices order.

If I got you right, you want not to find changes, but symmetric difference. For that, one approach might be concatenate dataframes:

>>> df = pd.concat([df1, df2])
>>> df = df.reset_index(drop=True)

group by

>>> df_gpby = df.groupby(list(df.columns))

get index of unique records

>>> idx = [x[0] for x in df_gpby.groups.values() if len(x) == 1]

filter

>>> df.reindex(idx)
         Date   Fruit   Num   Color
9  2013-11-25  Orange   8.6  Orange
8  2013-11-25   Apple  22.1     Red

解决方案 2：

Updating and placing, somewhere it will be easier for others to find, ling's comment upon jur's response above.

df_diff = pd.concat([df1,df2]).drop_duplicates(keep=False)

Testing with these DataFrames:

# with import pandas as pd

df1 = pd.DataFrame({
    'Date':['2013-11-24','2013-11-24','2013-11-24','2013-11-24'],
    'Fruit':['Banana','Orange','Apple','Celery'],
    'Num':[22.1,8.6,7.6,10.2],
    'Color':['Yellow','Orange','Green','Green'],
    })

df2 = pd.DataFrame({
    'Date':['2013-11-24','2013-11-24','2013-11-24','2013-11-24','2013-11-25','2013-11-25'],
    'Fruit':['Banana','Orange','Apple','Celery','Apple','Orange'],
    'Num':[22.1,8.6,7.6,10.2,22.1,8.6],
    'Color':['Yellow','Orange','Green','Green','Red','Orange'],
    })

Results in this:

# for df1

         Date   Fruit   Num   Color
0  2013-11-24  Banana  22.1  Yellow
1  2013-11-24  Orange   8.6  Orange
2  2013-11-24   Apple   7.6   Green
3  2013-11-24  Celery  10.2   Green


# for df2

         Date   Fruit   Num   Color
0  2013-11-24  Banana  22.1  Yellow
1  2013-11-24  Orange   8.6  Orange
2  2013-11-24   Apple   7.6   Green
3  2013-11-24  Celery  10.2   Green
4  2013-11-25   Apple  22.1     Red
5  2013-11-25  Orange   8.6  Orange


# for df_diff

         Date   Fruit   Num   Color
4  2013-11-25   Apple  22.1     Red
5  2013-11-25  Orange   8.6  Orange

解决方案 3：

# THIS WORK FOR ME

# Get all diferent values
df3 = pd.merge(df1, df2, how='outer', indicator='Exist')
df3 = df3.loc[df3['Exist'] != 'both']


# If you like to filter by a common ID
df3  = pd.merge(df1, df2, on="Fruit", how='outer', indicator='Exist')
df3  = df3.loc[df3['Exist'] != 'both']

解决方案 4：

Passing the dataframes to concat in a dictionary, results in a multi-index dataframe from which you can easily delete the duplicates, which results in a multi-index dataframe with the differences between the dataframes:

import sys
if sys.version_info[0] < 3:
    from StringIO import StringIO
else:
    from io import StringIO
import pandas as pd

DF1 = StringIO("""Date       Fruit  Num  Color 
2013-11-24 Banana 22.1 Yellow
2013-11-24 Orange  8.6 Orange
2013-11-24 Apple   7.6 Green
2013-11-24 Celery 10.2 Green
""")
DF2 = StringIO("""Date       Fruit  Num  Color 
2013-11-24 Banana 22.1 Yellow
2013-11-24 Orange  8.6 Orange
2013-11-24 Apple   7.6 Green
2013-11-24 Celery 10.2 Green
2013-11-25 Apple  22.1 Red
2013-11-25 Orange  8.6 Orange""")


df1 = pd.read_table(DF1, sep='s+')
df2 = pd.read_table(DF2, sep='s+')
#%%
dfs_dictionary = {'DF1':df1,'DF2':df2}
df=pd.concat(dfs_dictionary)
df.drop_duplicates(keep=False)

Result:

             Date   Fruit   Num   Color
DF2 4  2013-11-25   Apple  22.1     Red
    5  2013-11-25  Orange   8.6  Orange

解决方案 5：

Since pandas >= 1.1.0 we have DataFrame.compare and Series.compare.

Note: the method can only compare identically-labeled DataFrame objects,
this means DataFrames with identical row and column labels.

df1 = pd.DataFrame({'A': [1, 2, 3],
                    'B': [4, 5, 6],
                    'C': [7, np.NaN, 9]})

df2 = pd.DataFrame({'A': [1, 99, 3],
                    'B': [4, 5, 81],
                    'C': [7, 8, 9]})

   A  B    C
0  1  4  7.0
1  2  5  NaN
2  3  6  9.0 

    A   B  C
0   1   4  7
1  99   5  8
2   3  81  9

df1.compare(df2)

     A          B          C      
  self other self other self other
1  2.0  99.0  NaN   NaN  NaN   8.0
2  NaN   NaN  6.0  81.0  NaN   NaN

解决方案 6：

Building on alko's answer that almost worked for me, except for the filtering step (where I get: ValueError: cannot reindex from a duplicate axis), here is the final solution I used:

# join the dataframes
united_data = pd.concat([data1, data2, data3, ...])
# group the data by the whole row to find duplicates
united_data_grouped = united_data.groupby(list(united_data.columns))
# detect the row indices of unique rows
uniq_data_idx = [x[0] for x in united_data_grouped.indices.values() if len(x) == 1]
# extract those unique values
uniq_data = united_data.iloc[uniq_data_idx]

解决方案 7：

Get the existing data from df2 into df1:

dfe = df2[df2["Fruit"].isin(df1["Fruit"])]

Get the non-existing data from df2 into df1:

dfn = df2[~ df2["Fruit"].isin(df1["Fruit"])]

You can use more than one comparison.

解决方案 8：

Founder a simple solution here:

https://stackoverflow.com/a/47132808/9656339

pd.concat([df1, df2]).loc[df1.index.symmetric_difference(df2.index)]

解决方案 9：

There is a simpler solution that is faster and better,
and if the numbers are different can even give you quantities differences:

df1_i = df1.set_index(['Date','Fruit','Color'])
df2_i = df2.set_index(['Date','Fruit','Color'])
df_diff = df1_i.join(df2_i,how='outer',rsuffix='_').fillna(0)
df_diff = (df_diff['Num'] - df_diff['Num_'])

Here df_diff is a synopsis of the differences. You can even use it to find the differences in quantities. In your example:

Explanation:
Similarly to comparing two lists, to do it efficiently we should first order them then compare them (converting the list to sets/hashing would also be fast; both are an incredible improvement to the simple O(N^2) double comparison loop

Note: the following code produces the tables:

df1=pd.DataFrame({
    'Date':['2013-11-24','2013-11-24','2013-11-24','2013-11-24'],
    'Fruit':['Banana','Orange','Apple','Celery'],
    'Num':[22.1,8.6,7.6,10.2],
    'Color':['Yellow','Orange','Green','Green'],
})
df2=pd.DataFrame({
    'Date':['2013-11-24','2013-11-24','2013-11-24','2013-11-24','2013-11-25','2013-11-25'],
    'Fruit':['Banana','Orange','Apple','Celery','Apple','Orange'],
    'Num':[22.1,8.6,7.6,10.2,22.1,8.6],
    'Color':['Yellow','Orange','Green','Green','Red','Orange'],
})

解决方案 10：

# given
df1=pd.DataFrame({'Date':['2013-11-24','2013-11-24','2013-11-24','2013-11-24'],
    'Fruit':['Banana','Orange','Apple','Celery'],
    'Num':[22.1,8.6,7.6,10.2],
    'Color':['Yellow','Orange','Green','Green']})
df2=pd.DataFrame({'Date':['2013-11-24','2013-11-24','2013-11-24','2013-11-24','2013-11-25','2013-11-25'],
    'Fruit':['Banana','Orange','Apple','Celery','Apple','Orange'],
    'Num':[22.1,8.6,7.6,1000,22.1,8.6],
    'Color':['Yellow','Orange','Green','Green','Red','Orange']})

# find which rows are in df2 that aren't in df1 by Date and Fruit
df_2notin1 = df2[~(df2['Date'].isin(df1['Date']) & df2['Fruit'].isin(df1['Fruit']) )].dropna().reset_index(drop=True)

# output
print('df_2notin1
', df_2notin1)
#      Color        Date   Fruit   Num
# 0     Red  2013-11-25   Apple  22.1
# 1  Orange  2013-11-25  Orange   8.6

解决方案 11：

I got this solution. Does this help you ?

text = """df1:
2013-11-24 Banana 22.1 Yellow
2013-11-24 Orange 8.6 Orange
2013-11-24 Apple 7.6 Green
2013-11-24 Celery 10.2 Green

df2:
2013-11-24 Banana 22.1 Yellow
2013-11-24 Orange 8.6 Orange
2013-11-24 Apple 7.6 Green
2013-11-24 Celery 10.2 Green
2013-11-25 Apple 22.1 Red
2013-11-25 Orange 8.6 Orange



argetz45
2013-11-24 Banana 22.1 Yellow
2013-11-24 Orange 118.6 Orange
2013-11-24 Apple 74.6 Green
2013-11-24 Celery 10.2 Green
2013-11-25     Nuts    45.8 Brown
2013-11-25 Apple 22.1 Red
2013-11-25 Orange 8.6 Orange
2013-11-26   Pear 102.54    Pale"""

.

from collections import OrderedDict
import re

r = re.compile('([a-zA-Zd]+).*
'
               '(20dd-[01]d-[0123]d.+
?'
               '(.+
?)*)'
               '(?=[ 
]*Z'
                  '|'
                  '
+[a-zA-Zd]+.*
'
                  '20dd-[01]d-[0123]d)')

r2 = re.compile('((20dd-[01]d-[0123]d) +([^d.]+)(?<! )[^
]+)')

d = OrderedDict()
bef = []

for m in r.finditer(text):
    li = []
    for x in r2.findall(m.group(2)):
        if not any(x[1:3]==elbef for elbef in bef):
            bef.append(x[1:3])
            li.append(x[0])
    d[m.group(1)] = li


for name,lu in d.iteritems():
    print '%s
%s
' % (name,'
'.join(lu))

result

df1
2013-11-24 Banana 22.1 Yellow
2013-11-24 Orange 8.6 Orange
2013-11-24 Apple 7.6 Green
2013-11-24 Celery 10.2 Green

df2
2013-11-25 Apple 22.1 Red
2013-11-25 Orange 8.6 Orange

argetz45
2013-11-25     Nuts    45.8 Brown
2013-11-26   Pear 102.54    Pale

解决方案 12：

I tried this method, and it worked. I hope it can help too:

"""Identify differences between two pandas DataFrames"""
df1.sort_index(inplace=True)
df2.sort_index(inplace=True)
df_all = pd.concat([df1, df12], axis='columns', keys=['First', 'Second'])
df_final = df_all.swaplevel(axis='columns')[df1.columns[1:]]
df_final[df_final['change this to one of the columns'] != df_final['change this to one of the columns']]

解决方案 13：

use merge outer to find the left outer values whose value is null

txt1="""Date,Fruit,Num,Color 
2013-11-24,Banana,22.1,Yellow
2013-11-24,Orange,8.6,Orange
2013-11-24,Apple,7.6,Green
2013-11-24,Celery,10.2,Green"""

txt2="""Date,Fruit,Num,Color 
2013-11-24,Banana,22.1,Yellow
2013-11-24,Orange,8.6,Orange
2013-11-24,Apple,7.6,Green
2013-11-24,Celery,10.2,Green
2013-11-25,Apple,22.1,Red
2013-11-25,Orange,8.6,Orange"""

from io import StringIO
f = StringIO(txt1)
df1 = pd.read_table(f,sep =',')
df1.set_index('Date',inplace=True)

f = StringIO(txt2)
df2 = pd.read_table(f,sep =',')
df2.set_index('Date',inplace=True)

df3 =pd.merge(df2, df1, left_index=True, right_index=True,  how='outer', 
     indicator=True
         ,suffixes=("", "_left")
         ).query("_merge=='left_only'")
remove_columns=[item for item in df3.columns if '_left' in item]
remove_columns.append('_merge')
df3=df3.drop(columns=remove_columns)
print(df3)

output:

         Date   Fruit   Num  Color 
0  2013-11-25   Apple  22.1     Red
1  2013-11-25  Orange   8.6  Orange

解决方案 14：

One important detail to notice is that your data has duplicate index values, so to perform any straightforward comparison we need to turn everything as unique with df.reset_index() and therefore we can perform selections based on conditions. Once in your case the index is defined, I assume that you would like to keep de index so there are a one-line solution:

[~df2.reset_index().isin(df1.reset_index())].dropna().set_index('Date')

Once the objective from a pythonic perspective is to improve readability, we can break a little bit:

# keep the index name, if it does not have a name it uses the default name
index_name = df.index.name if df.index.name else 'index' 

# setting the index to become unique
df1 = df1.reset_index()
df2 = df2.reset_index()

# getting the differences to a Dataframe
df_diff = df2[~df2.isin(df1)].dropna().set_index(index_name)

解决方案 15：

Hope this would be useful to you. ^o^

df1 = pd.DataFrame({'date': ['0207', '0207'], 'col1': [1, 2]})
df2 = pd.DataFrame({'date': ['0207', '0207', '0208', '0208'], 'col1': [1, 2, 3, 4]})
print(f"df1(Before):
{df1}
df2:
{df2}")
"""
df1(Before):
   date  col1
0  0207     1
1  0207     2

df2:
   date  col1
0  0207     1
1  0207     2
2  0208     3
3  0208     4
"""

old_set = set(df1.index.values)
new_set = set(df2.index.values)
new_data_index = new_set - old_set
new_data_list = []
for idx in new_data_index:
    new_data_list.append(df2.loc[idx])

if len(new_data_list) > 0:
    df1 = df1.append(new_data_list)
print(f"df1(After):
{df1}")
"""
df1(After):
   date  col1
0  0207     1
1  0207     2
2  0208     3
3  0208     4
"""

解决方案 16：

You can find the difference between DataFrame row counts:

df2.value_counts().sub(df1.value_counts(), fill_value=0)

Output:

Date        Fruit   Num     Color
2013-11-24  Apple   7.6     Green     0.0
            Banana  22.1    Yellow    0.0
            Celery  10.2    Green    -1.0
                    1000.0  Green     1.0
            Orange  8.6     Orange    0.0
2013-11-25  Apple   22.1    Red       1.0
            Orange  8.6     Orange    1.0
dtype: float6