标签归档:panel

比较两个DataFrame并并排输出它们的差异

问题:比较两个DataFrame并并排输出它们的差异

我试图突出显示两个数据框之间到底发生了什么变化。

假设我有两个Python Pandas数据框:

"StudentRoster Jan-1":
id   Name   score                    isEnrolled           Comment
111  Jack   2.17                     True                 He was late to class
112  Nick   1.11                     False                Graduated
113  Zoe    4.12                     True       

"StudentRoster Jan-2":
id   Name   score                    isEnrolled           Comment
111  Jack   2.17                     True                 He was late to class
112  Nick   1.21                     False                Graduated
113  Zoe    4.12                     False                On vacation

我的目标是输出一个HTML表:

  1. 标识已更改的行(可以是int,float,boolean,string)
  2. 输出具有相同,OLD和NEW值的行(理想情况下将其输出到HTML表中),以便使用者可以清楚地看到两个数据框之间的变化:

    "StudentRoster Difference Jan-1 - Jan-2":  
    id   Name   score                    isEnrolled           Comment
    112  Nick   was 1.11| now 1.21       False                Graduated
    113  Zoe    4.12                     was True | now False was "" | now   "On   vacation"

我想我可以逐行和逐列进行比较,但是有没有更简单的方法?

I am trying to highlight exactly what changed between two dataframes.

Suppose I have two Python Pandas dataframes:

"StudentRoster Jan-1":
id   Name   score                    isEnrolled           Comment
111  Jack   2.17                     True                 He was late to class
112  Nick   1.11                     False                Graduated
113  Zoe    4.12                     True       

"StudentRoster Jan-2":
id   Name   score                    isEnrolled           Comment
111  Jack   2.17                     True                 He was late to class
112  Nick   1.21                     False                Graduated
113  Zoe    4.12                     False                On vacation

My goal is to output an HTML table that:

  1. Identifies rows that have changed (could be int, float, boolean, string)
  2. Outputs rows with same, OLD and NEW values (ideally into an HTML table) so the consumer can clearly see what changed between two dataframes:

    "StudentRoster Difference Jan-1 - Jan-2":  
    id   Name   score                    isEnrolled           Comment
    112  Nick   was 1.11| now 1.21       False                Graduated
    113  Zoe    4.12                     was True | now False was "" | now   "On   vacation"
    

I suppose I could do a row by row and column by column comparison, but is there an easier way?


回答 0

第一部分类似于君士坦丁,您可以获取哪些行为空的布尔值*:

In [21]: ne = (df1 != df2).any(1)

In [22]: ne
Out[22]:
0    False
1     True
2     True
dtype: bool

然后,我们可以查看哪些条目已更改:

In [23]: ne_stacked = (df1 != df2).stack()

In [24]: changed = ne_stacked[ne_stacked]

In [25]: changed.index.names = ['id', 'col']

In [26]: changed
Out[26]:
id  col
1   score         True
2   isEnrolled    True
    Comment       True
dtype: bool

在这里,第一个条目是索引,第二个条目是已更改的列。

In [27]: difference_locations = np.where(df1 != df2)

In [28]: changed_from = df1.values[difference_locations]

In [29]: changed_to = df2.values[difference_locations]

In [30]: pd.DataFrame({'from': changed_from, 'to': changed_to}, index=changed.index)
Out[30]:
               from           to
id col
1  score       1.11         1.21
2  isEnrolled  True        False
   Comment     None  On vacation

*注:这是非常重要的df1,并df2在这里分享相同的索引。为了克服这种歧义,您可以确保仅使用来查看共享标签df1.index & df2.index,但我想将其保留为练习。

The first part is similar to Constantine, you can get the boolean of which rows are empty*:

In [21]: ne = (df1 != df2).any(1)

In [22]: ne
Out[22]:
0    False
1     True
2     True
dtype: bool

Then we can see which entries have changed:

In [23]: ne_stacked = (df1 != df2).stack()

In [24]: changed = ne_stacked[ne_stacked]

In [25]: changed.index.names = ['id', 'col']

In [26]: changed
Out[26]:
id  col
1   score         True
2   isEnrolled    True
    Comment       True
dtype: bool

Here the first entry is the index and the second the columns which has been changed.

In [27]: difference_locations = np.where(df1 != df2)

In [28]: changed_from = df1.values[difference_locations]

In [29]: changed_to = df2.values[difference_locations]

In [30]: pd.DataFrame({'from': changed_from, 'to': changed_to}, index=changed.index)
Out[30]:
               from           to
id col
1  score       1.11         1.21
2  isEnrolled  True        False
   Comment     None  On vacation

* Note: it’s important that df1 and df2 share the same index here. To overcome this ambiguity, you can ensure you only look at the shared labels using df1.index & df2.index, but I think I’ll leave that as an exercise.


回答 1

突出显示两个DataFrame之间的差异

可以使用DataFrame样式属性突出显示存在差异的单元格的背景色。

使用原始问题中的示例数据

第一步是使用concat功能将DataFrames水平连接,并使用keys参数区分每个帧:

df_all = pd.concat([df.set_index('id'), df2.set_index('id')], 
                   axis='columns', keys=['First', 'Second'])
df_all

交换列级别并将相同的列名称彼此相邻可能更容易:

df_final = df_all.swaplevel(axis='columns')[df.columns[1:]]
df_final

现在,更容易发现框架中的差异。但是,我们可以走得更远,并使用该style属性突出显示不同的单元格。我们定义了一个自定义函数来执行此操作,您可以在文档的此部分中看到。

def highlight_diff(data, color='yellow'):
    attr = 'background-color: {}'.format(color)
    other = data.xs('First', axis='columns', level=-1)
    return pd.DataFrame(np.where(data.ne(other, level=0), attr, ''),
                        index=data.index, columns=data.columns)

df_final.style.apply(highlight_diff, axis=None)

这将突出显示两个均缺少值的单元格。您可以填充它们或提供额外的逻辑,以免突出显示它们。

Highlighting the difference between two DataFrames

It is possible to use the DataFrame style property to highlight the background color of the cells where there is a difference.

Using the example data from the original question

The first step is to concatenate the DataFrames horizontally with the concat function and distinguish each frame with the keys parameter:

df_all = pd.concat([df.set_index('id'), df2.set_index('id')], 
                   axis='columns', keys=['First', 'Second'])
df_all

It’s probably easier to swap the column levels and put the same column names next to each other:

df_final = df_all.swaplevel(axis='columns')[df.columns[1:]]
df_final

Now, its much easier to spot the differences in the frames. But, we can go further and use the style property to highlight the cells that are different. We define a custom function to do this which you can see in this part of the documentation.

def highlight_diff(data, color='yellow'):
    attr = 'background-color: {}'.format(color)
    other = data.xs('First', axis='columns', level=-1)
    return pd.DataFrame(np.where(data.ne(other, level=0), attr, ''),
                        index=data.index, columns=data.columns)

df_final.style.apply(highlight_diff, axis=None)

This will highlight cells that both have missing values. You can either fill them or provide extra logic so that they don’t get highlighted.


回答 2

这个答案只是扩展了@Andy Hayden的值,使其在数字字段为时具有弹性nan,并将其包装到函数中。

import pandas as pd
import numpy as np


def diff_pd(df1, df2):
    """Identify differences between two pandas DataFrames"""
    assert (df1.columns == df2.columns).all(), \
        "DataFrame column names are different"
    if any(df1.dtypes != df2.dtypes):
        "Data Types are different, trying to convert"
        df2 = df2.astype(df1.dtypes)
    if df1.equals(df2):
        return None
    else:
        # need to account for np.nan != np.nan returning True
        diff_mask = (df1 != df2) & ~(df1.isnull() & df2.isnull())
        ne_stacked = diff_mask.stack()
        changed = ne_stacked[ne_stacked]
        changed.index.names = ['id', 'col']
        difference_locations = np.where(diff_mask)
        changed_from = df1.values[difference_locations]
        changed_to = df2.values[difference_locations]
        return pd.DataFrame({'from': changed_from, 'to': changed_to},
                            index=changed.index)

因此,对于您的数据(略作编辑以使分数列中具有NaN):

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

DF1 = StringIO("""id   Name   score                    isEnrolled           Comment
111  Jack   2.17                     True                 "He was late to class"
112  Nick   1.11                     False                "Graduated"
113  Zoe    NaN                     True                  " "
""")
DF2 = StringIO("""id   Name   score                    isEnrolled           Comment
111  Jack   2.17                     True                 "He was late to class"
112  Nick   1.21                     False                "Graduated"
113  Zoe    NaN                     False                "On vacation" """)
df1 = pd.read_table(DF1, sep='\s+', index_col='id')
df2 = pd.read_table(DF2, sep='\s+', index_col='id')
diff_pd(df1, df2)

输出:

                from           to
id  col                          
112 score       1.11         1.21
113 isEnrolled  True        False
    Comment           On vacation

This answer simply extends @Andy Hayden’s, making it resilient to when numeric fields are nan, and wrapping it up into a function.

import pandas as pd
import numpy as np


def diff_pd(df1, df2):
    """Identify differences between two pandas DataFrames"""
    assert (df1.columns == df2.columns).all(), \
        "DataFrame column names are different"
    if any(df1.dtypes != df2.dtypes):
        "Data Types are different, trying to convert"
        df2 = df2.astype(df1.dtypes)
    if df1.equals(df2):
        return None
    else:
        # need to account for np.nan != np.nan returning True
        diff_mask = (df1 != df2) & ~(df1.isnull() & df2.isnull())
        ne_stacked = diff_mask.stack()
        changed = ne_stacked[ne_stacked]
        changed.index.names = ['id', 'col']
        difference_locations = np.where(diff_mask)
        changed_from = df1.values[difference_locations]
        changed_to = df2.values[difference_locations]
        return pd.DataFrame({'from': changed_from, 'to': changed_to},
                            index=changed.index)

So with your data (slightly edited to have a NaN in the score column):

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

DF1 = StringIO("""id   Name   score                    isEnrolled           Comment
111  Jack   2.17                     True                 "He was late to class"
112  Nick   1.11                     False                "Graduated"
113  Zoe    NaN                     True                  " "
""")
DF2 = StringIO("""id   Name   score                    isEnrolled           Comment
111  Jack   2.17                     True                 "He was late to class"
112  Nick   1.21                     False                "Graduated"
113  Zoe    NaN                     False                "On vacation" """)
df1 = pd.read_table(DF1, sep='\s+', index_col='id')
df2 = pd.read_table(DF2, sep='\s+', index_col='id')
diff_pd(df1, df2)

Output:

                from           to
id  col                          
112 score       1.11         1.21
113 isEnrolled  True        False
    Comment           On vacation

回答 3

import pandas as pd
import io

texts = ['''\
id   Name   score                    isEnrolled                        Comment
111  Jack   2.17                     True                 He was late to class
112  Nick   1.11                     False                           Graduated
113  Zoe    4.12                     True       ''',

         '''\
id   Name   score                    isEnrolled                        Comment
111  Jack   2.17                     True                 He was late to class
112  Nick   1.21                     False                           Graduated
113  Zoe    4.12                     False                         On vacation''']


df1 = pd.read_fwf(io.StringIO(texts[0]), widths=[5,7,25,21,20])
df2 = pd.read_fwf(io.StringIO(texts[1]), widths=[5,7,25,21,20])
df = pd.concat([df1,df2]) 

print(df)
#     id  Name  score isEnrolled               Comment
# 0  111  Jack   2.17       True  He was late to class
# 1  112  Nick   1.11      False             Graduated
# 2  113   Zoe   4.12       True                   NaN
# 0  111  Jack   2.17       True  He was late to class
# 1  112  Nick   1.21      False             Graduated
# 2  113   Zoe   4.12      False           On vacation

df.set_index(['id', 'Name'], inplace=True)
print(df)
#           score isEnrolled               Comment
# id  Name                                        
# 111 Jack   2.17       True  He was late to class
# 112 Nick   1.11      False             Graduated
# 113 Zoe    4.12       True                   NaN
# 111 Jack   2.17       True  He was late to class
# 112 Nick   1.21      False             Graduated
# 113 Zoe    4.12      False           On vacation

def report_diff(x):
    return x[0] if x[0] == x[1] else '{} | {}'.format(*x)

changes = df.groupby(level=['id', 'Name']).agg(report_diff)
print(changes)

版画

                score    isEnrolled               Comment
id  Name                                                 
111 Jack         2.17          True  He was late to class
112 Nick  1.11 | 1.21         False             Graduated
113 Zoe          4.12  True | False     nan | On vacation
import pandas as pd
import io

texts = ['''\
id   Name   score                    isEnrolled                        Comment
111  Jack   2.17                     True                 He was late to class
112  Nick   1.11                     False                           Graduated
113  Zoe    4.12                     True       ''',

         '''\
id   Name   score                    isEnrolled                        Comment
111  Jack   2.17                     True                 He was late to class
112  Nick   1.21                     False                           Graduated
113  Zoe    4.12                     False                         On vacation''']


df1 = pd.read_fwf(io.StringIO(texts[0]), widths=[5,7,25,21,20])
df2 = pd.read_fwf(io.StringIO(texts[1]), widths=[5,7,25,21,20])
df = pd.concat([df1,df2]) 

print(df)
#     id  Name  score isEnrolled               Comment
# 0  111  Jack   2.17       True  He was late to class
# 1  112  Nick   1.11      False             Graduated
# 2  113   Zoe   4.12       True                   NaN
# 0  111  Jack   2.17       True  He was late to class
# 1  112  Nick   1.21      False             Graduated
# 2  113   Zoe   4.12      False           On vacation

df.set_index(['id', 'Name'], inplace=True)
print(df)
#           score isEnrolled               Comment
# id  Name                                        
# 111 Jack   2.17       True  He was late to class
# 112 Nick   1.11      False             Graduated
# 113 Zoe    4.12       True                   NaN
# 111 Jack   2.17       True  He was late to class
# 112 Nick   1.21      False             Graduated
# 113 Zoe    4.12      False           On vacation

def report_diff(x):
    return x[0] if x[0] == x[1] else '{} | {}'.format(*x)

changes = df.groupby(level=['id', 'Name']).agg(report_diff)
print(changes)

prints

                score    isEnrolled               Comment
id  Name                                                 
111 Jack         2.17          True  He was late to class
112 Nick  1.11 | 1.21         False             Graduated
113 Zoe          4.12  True | False     nan | On vacation

回答 4

我已经遇到了这个问题,但是在找到这篇文章之前找到了答案:

根据unutbu的答案,加载您的数据…

import pandas as pd
import io

texts = ['''\
id   Name   score                    isEnrolled                       Date
111  Jack                            True              2013-05-01 12:00:00
112  Nick   1.11                     False             2013-05-12 15:05:23
     Zoe    4.12                     True                                  ''',

         '''\
id   Name   score                    isEnrolled                       Date
111  Jack   2.17                     True              2013-05-01 12:00:00
112  Nick   1.21                     False                                
     Zoe    4.12                     False             2013-05-01 12:00:00''']


df1 = pd.read_fwf(io.StringIO(texts[0]), widths=[5,7,25,17,20], parse_dates=[4])
df2 = pd.read_fwf(io.StringIO(texts[1]), widths=[5,7,25,17,20], parse_dates=[4])

…定义您的diff函数…

def report_diff(x):
    return x[0] if x[0] == x[1] else '{} | {}'.format(*x)

然后,您可以简单地使用面板来得出结论:

my_panel = pd.Panel(dict(df1=df1,df2=df2))
print my_panel.apply(report_diff, axis=0)

#          id  Name        score    isEnrolled                       Date
#0        111  Jack   nan | 2.17          True        2013-05-01 12:00:00
#1        112  Nick  1.11 | 1.21         False  2013-05-12 15:05:23 | NaT
#2  nan | nan   Zoe         4.12  True | False  NaT | 2013-05-01 12:00:00

顺便说一句,如果您使用的是IPython Notebook,则可能希望使用彩色的diff函数根据单元格是不同,相等还是left / right null来赋予颜色:

from IPython.display import HTML
pd.options.display.max_colwidth = 500  # You need this, otherwise pandas
#                          will limit your HTML strings to 50 characters

def report_diff(x):
    if x[0]==x[1]:
        return unicode(x[0].__str__())
    elif pd.isnull(x[0]) and pd.isnull(x[1]):
        return u'<table style="background-color:#00ff00;font-weight:bold;">'+\
            '<tr><td>%s</td></tr><tr><td>%s</td></tr></table>' % ('nan', 'nan')
    elif pd.isnull(x[0]) and ~pd.isnull(x[1]):
        return u'<table style="background-color:#ffff00;font-weight:bold;">'+\
            '<tr><td>%s</td></tr><tr><td>%s</td></tr></table>' % ('nan', x[1])
    elif ~pd.isnull(x[0]) and pd.isnull(x[1]):
        return u'<table style="background-color:#0000ff;font-weight:bold;">'+\
            '<tr><td>%s</td></tr><tr><td>%s</td></tr></table>' % (x[0],'nan')
    else:
        return u'<table style="background-color:#ff0000;font-weight:bold;">'+\
            '<tr><td>%s</td></tr><tr><td>%s</td></tr></table>' % (x[0], x[1])

HTML(my_panel.apply(report_diff, axis=0).to_html(escape=False))

I have faced this issue, but found an answer before finding this post :

Based on unutbu’s answer, load your data…

import pandas as pd
import io

texts = ['''\
id   Name   score                    isEnrolled                       Date
111  Jack                            True              2013-05-01 12:00:00
112  Nick   1.11                     False             2013-05-12 15:05:23
     Zoe    4.12                     True                                  ''',

         '''\
id   Name   score                    isEnrolled                       Date
111  Jack   2.17                     True              2013-05-01 12:00:00
112  Nick   1.21                     False                                
     Zoe    4.12                     False             2013-05-01 12:00:00''']


df1 = pd.read_fwf(io.StringIO(texts[0]), widths=[5,7,25,17,20], parse_dates=[4])
df2 = pd.read_fwf(io.StringIO(texts[1]), widths=[5,7,25,17,20], parse_dates=[4])

…define your diff function…

def report_diff(x):
    return x[0] if x[0] == x[1] else '{} | {}'.format(*x)

Then you can simply use a Panel to conclude :

my_panel = pd.Panel(dict(df1=df1,df2=df2))
print my_panel.apply(report_diff, axis=0)

#          id  Name        score    isEnrolled                       Date
#0        111  Jack   nan | 2.17          True        2013-05-01 12:00:00
#1        112  Nick  1.11 | 1.21         False  2013-05-12 15:05:23 | NaT
#2  nan | nan   Zoe         4.12  True | False  NaT | 2013-05-01 12:00:00

By the way, if you’re in IPython Notebook, you may like to use a colored diff function to give colors depending whether cells are different, equal or left/right null :

from IPython.display import HTML
pd.options.display.max_colwidth = 500  # You need this, otherwise pandas
#                          will limit your HTML strings to 50 characters

def report_diff(x):
    if x[0]==x[1]:
        return unicode(x[0].__str__())
    elif pd.isnull(x[0]) and pd.isnull(x[1]):
        return u'<table style="background-color:#00ff00;font-weight:bold;">'+\
            '<tr><td>%s</td></tr><tr><td>%s</td></tr></table>' % ('nan', 'nan')
    elif pd.isnull(x[0]) and ~pd.isnull(x[1]):
        return u'<table style="background-color:#ffff00;font-weight:bold;">'+\
            '<tr><td>%s</td></tr><tr><td>%s</td></tr></table>' % ('nan', x[1])
    elif ~pd.isnull(x[0]) and pd.isnull(x[1]):
        return u'<table style="background-color:#0000ff;font-weight:bold;">'+\
            '<tr><td>%s</td></tr><tr><td>%s</td></tr></table>' % (x[0],'nan')
    else:
        return u'<table style="background-color:#ff0000;font-weight:bold;">'+\
            '<tr><td>%s</td></tr><tr><td>%s</td></tr></table>' % (x[0], x[1])

HTML(my_panel.apply(report_diff, axis=0).to_html(escape=False))

回答 5

如果您的两个数据帧中具有相同的ID,那么找出更改实际上是很容易的。这样做frame1 != frame2会为您提供一个布尔型DataFrame,其中每个True都是已更改的数据。由此,您可以通过轻松获得每个更改行的索引changedids = frame1.index[np.any(frame1 != frame2,axis=1)]

If your two dataframes have the same ids in them, then finding out what changed is actually pretty easy. Just doing frame1 != frame2 will give you a boolean DataFrame where each True is data that has changed. From that, you could easily get the index of each changed row by doing changedids = frame1.index[np.any(frame1 != frame2,axis=1)].


回答 6

使用concat和drop_duplicates的另一种方法:

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

DF1 = StringIO("""id   Name   score                    isEnrolled           Comment
111  Jack   2.17                     True                 "He was late to class"
112  Nick   1.11                     False                "Graduated"
113  Zoe    NaN                     True                  " "
""")
DF2 = StringIO("""id   Name   score                    isEnrolled           Comment
111  Jack   2.17                     True                 "He was late to class"
112  Nick   1.21                     False                "Graduated"
113  Zoe    NaN                     False                "On vacation" """)

df1 = pd.read_table(DF1, sep='\s+', index_col='id')
df2 = pd.read_table(DF2, sep='\s+', index_col='id')
#%%
dictionary = {1:df1,2:df2}
df=pd.concat(dictionary)
df.drop_duplicates(keep=False)

输出:

       Name  score isEnrolled      Comment
  id                                      
1 112  Nick   1.11      False    Graduated
  113   Zoe    NaN       True             
2 112  Nick   1.21      False    Graduated
  113   Zoe    NaN      False  On vacation

A different approach using concat and drop_duplicates:

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

DF1 = StringIO("""id   Name   score                    isEnrolled           Comment
111  Jack   2.17                     True                 "He was late to class"
112  Nick   1.11                     False                "Graduated"
113  Zoe    NaN                     True                  " "
""")
DF2 = StringIO("""id   Name   score                    isEnrolled           Comment
111  Jack   2.17                     True                 "He was late to class"
112  Nick   1.21                     False                "Graduated"
113  Zoe    NaN                     False                "On vacation" """)

df1 = pd.read_table(DF1, sep='\s+', index_col='id')
df2 = pd.read_table(DF2, sep='\s+', index_col='id')
#%%
dictionary = {1:df1,2:df2}
df=pd.concat(dictionary)
df.drop_duplicates(keep=False)

Output:

       Name  score isEnrolled      Comment
  id                                      
1 112  Nick   1.11      False    Graduated
  113   Zoe    NaN       True             
2 112  Nick   1.21      False    Graduated
  113   Zoe    NaN      False  On vacation

回答 7

摆弄@journois的答案后,由于Panel的贬值,我能够使用MultiIndex而不是Panel使它正常工作

首先,创建一些虚拟数据:

df1 = pd.DataFrame({
    'id': ['111', '222', '333', '444', '555'],
    'let': ['a', 'b', 'c', 'd', 'e'],
    'num': ['1', '2', '3', '4', '5']
})
df2 = pd.DataFrame({
    'id': ['111', '222', '333', '444', '666'],
    'let': ['a', 'b', 'c', 'D', 'f'],
    'num': ['1', '2', 'Three', '4', '6'],
})

然后,定义您的diff函数,在这种情况下,我将使用他的答案中的一个report_diff保持不变:

def report_diff(x):
    return x[0] if x[0] == x[1] else '{} | {}'.format(*x)

然后,我将把数据连接到一个MultiIndex数据帧中:

df_all = pd.concat(
    [df1.set_index('id'), df2.set_index('id')], 
    axis='columns', 
    keys=['df1', 'df2'],
    join='outer'
)
df_all = df_all.swaplevel(axis='columns')[df1.columns[1:]]

最后,我将report_diff向下应用每个列组:

df_final.groupby(level=0, axis=1).apply(lambda frame: frame.apply(report_diff, axis=1))

输出:

         let        num
111        a          1
222        b          2
333        c  3 | Three
444    d | D          4
555  e | nan    5 | nan
666  nan | f    nan | 6

仅此而已!

After fiddling around with @journois’s answer, I was able to get it to work using MultiIndex instead of Panel due to Panel’s deprication.

First, create some dummy data:

df1 = pd.DataFrame({
    'id': ['111', '222', '333', '444', '555'],
    'let': ['a', 'b', 'c', 'd', 'e'],
    'num': ['1', '2', '3', '4', '5']
})
df2 = pd.DataFrame({
    'id': ['111', '222', '333', '444', '666'],
    'let': ['a', 'b', 'c', 'D', 'f'],
    'num': ['1', '2', 'Three', '4', '6'],
})

Then, define your diff function, in this case I’ll use the one from his answer report_diff stays the same:

def report_diff(x):
    return x[0] if x[0] == x[1] else '{} | {}'.format(*x)

Then, I’m going to concatenate the data into a MultiIndex dataframe:

df_all = pd.concat(
    [df1.set_index('id'), df2.set_index('id')], 
    axis='columns', 
    keys=['df1', 'df2'],
    join='outer'
)
df_all = df_all.swaplevel(axis='columns')[df1.columns[1:]]

And finally I’m going to apply the report_diff down each column group:

df_final.groupby(level=0, axis=1).apply(lambda frame: frame.apply(report_diff, axis=1))

This outputs:

         let        num
111        a          1
222        b          2
333        c  3 | Three
444    d | D          4
555  e | nan    5 | nan
666  nan | f    nan | 6

And that is all!


回答 8

扩展@cge的答案,这对于提高结果的可读性非常酷:

a[a != b][np.any(a != b, axis=1)].join(pd.DataFrame('a<->b', index=a.index, columns=['a<=>b'])).join(
        b[a != b][np.any(a != b, axis=1)]
        ,rsuffix='_b', how='outer'
).fillna('')

完整的演示示例:

import numpy as np, pandas as pd

a = pd.DataFrame(np.random.randn(7,3), columns=list('ABC'))
b = a.copy()
b.iloc[0,2] = np.nan
b.iloc[1,0] = 7
b.iloc[3,1] = 77
b.iloc[4,2] = 777

a[a != b][np.any(a != b, axis=1)].join(pd.DataFrame('a<->b', index=a.index, columns=['a<=>b'])).join(
        b[a != b][np.any(a != b, axis=1)]
        ,rsuffix='_b', how='outer'
).fillna('')

Extending answer of @cge, which is pretty cool for more readability of result:

a[a != b][np.any(a != b, axis=1)].join(pd.DataFrame('a<->b', index=a.index, columns=['a<=>b'])).join(
        b[a != b][np.any(a != b, axis=1)]
        ,rsuffix='_b', how='outer'
).fillna('')

Full demonstration example:

import numpy as np, pandas as pd

a = pd.DataFrame(np.random.randn(7,3), columns=list('ABC'))
b = a.copy()
b.iloc[0,2] = np.nan
b.iloc[1,0] = 7
b.iloc[3,1] = 77
b.iloc[4,2] = 777

a[a != b][np.any(a != b, axis=1)].join(pd.DataFrame('a<->b', index=a.index, columns=['a<=>b'])).join(
        b[a != b][np.any(a != b, axis=1)]
        ,rsuffix='_b', how='outer'
).fillna('')

回答 9

这是使用选择并合并的另一种方法:

In [6]: # first lets create some dummy dataframes with some column(s) different
   ...: df1 = pd.DataFrame({'a': range(-5,0), 'b': range(10,15), 'c': range(20,25)})
   ...: df2 = pd.DataFrame({'a': range(-5,0), 'b': range(10,15), 'c': [20] + list(range(101,105))})


In [7]: df1
Out[7]:
   a   b   c
0 -5  10  20
1 -4  11  21
2 -3  12  22
3 -2  13  23
4 -1  14  24


In [8]: df2
Out[8]:
   a   b    c
0 -5  10   20
1 -4  11  101
2 -3  12  102
3 -2  13  103
4 -1  14  104


In [10]: # make condition over the columns you want to comapre
    ...: condition = df1['c'] != df2['c']
    ...:
    ...: # select rows from each dataframe where the condition holds
    ...: diff1 = df1[condition]
    ...: diff2 = df2[condition]


In [11]: # merge the selected rows (dataframes) with some suffixes (optional)
    ...: diff1.merge(diff2, on=['a','b'], suffixes=('_before', '_after'))
Out[11]:
   a   b  c_before  c_after
0 -4  11        21      101
1 -3  12        22      102
2 -2  13        23      103
3 -1  14        24      104

这是Jupyter屏幕截图中的相同内容:

Here is another way using select and merge:

In [6]: # first lets create some dummy dataframes with some column(s) different
   ...: df1 = pd.DataFrame({'a': range(-5,0), 'b': range(10,15), 'c': range(20,25)})
   ...: df2 = pd.DataFrame({'a': range(-5,0), 'b': range(10,15), 'c': [20] + list(range(101,105))})


In [7]: df1
Out[7]:
   a   b   c
0 -5  10  20
1 -4  11  21
2 -3  12  22
3 -2  13  23
4 -1  14  24


In [8]: df2
Out[8]:
   a   b    c
0 -5  10   20
1 -4  11  101
2 -3  12  102
3 -2  13  103
4 -1  14  104


In [10]: # make condition over the columns you want to comapre
    ...: condition = df1['c'] != df2['c']
    ...:
    ...: # select rows from each dataframe where the condition holds
    ...: diff1 = df1[condition]
    ...: diff2 = df2[condition]


In [11]: # merge the selected rows (dataframes) with some suffixes (optional)
    ...: diff1.merge(diff2, on=['a','b'], suffixes=('_before', '_after'))
Out[11]:
   a   b  c_before  c_after
0 -4  11        21      101
1 -3  12        22      102
2 -2  13        23      103
3 -1  14        24      104

Here is the same thing from a Jupyter screenshot:


回答 10

大熊猫> = 1.1: DataFrame.compare

使用pandas 1.1,您基本上可以通过一个函数调用来复制Ted Petrou的输出。来自文档的示例:

pd.__version__
# '1.1.0.dev0+2004.g8d10bfb6f'

df1.compare(df2)

  score       isEnrolled       Comment             
   self other       self other    self        other
1  1.11  1.21        NaN   NaN     NaN          NaN
2   NaN   NaN        1.0   0.0     NaN  On vacation

此处,“自身”是指LHS数据帧,而“其他”是指RHS数据帧。默认情况下,相等值将替换为NaN,因此您可以仅关注差异。如果要显示相等的值,请使用

df1.compare(df2, keep_equal=True, keep_shape=True) 

  score       isEnrolled           Comment             
   self other       self  other       self        other
1  1.11  1.21      False  False  Graduated    Graduated
2  4.12  4.12       True  False        NaN  On vacation

您还可以使用align_axis以下方式更改比较轴:

df1.compare(df2, align_axis='index')

         score  isEnrolled      Comment
1 self    1.11         NaN          NaN
  other   1.21         NaN          NaN
2 self     NaN         1.0          NaN
  other    NaN         0.0  On vacation

这将按行而不是按列比较值。

pandas >= 1.1: DataFrame.compare

With pandas 1.1, you could essentially replicate Ted Petrou’s output with a single function call. Example taken from the docs:

pd.__version__
# '1.1.0'

df1.compare(df2)

  score       isEnrolled       Comment             
   self other       self other    self        other
1  1.11  1.21        NaN   NaN     NaN          NaN
2   NaN   NaN        1.0   0.0     NaN  On vacation

Here, “self” refers to the LHS dataFrame, while “other” is the RHS DataFrame. By default, equal values are replaced with NaNs so you can focus on just the diffs. If you want to show values that are equal as well, use

df1.compare(df2, keep_equal=True, keep_shape=True) 

  score       isEnrolled           Comment             
   self other       self  other       self        other
1  1.11  1.21      False  False  Graduated    Graduated
2  4.12  4.12       True  False        NaN  On vacation

You can also change the axis of comparison using align_axis:

df1.compare(df2, align_axis='index')

         score  isEnrolled      Comment
1 self    1.11         NaN          NaN
  other   1.21         NaN          NaN
2 self     NaN         1.0          NaN
  other    NaN         0.0  On vacation

This compares values row-wise, instead of column-wise.


回答 11

查找两个数据帧之间不对称差异的函数在以下实现:(基于熊猫的集合差异)GIST:https : //gist.github.com/oneryalcin/68cf25f536a25e65f0b3c84f9c118e03

def diff_df(df1, df2, how="left"):
    """
      Find Difference of rows for given two dataframes
      this function is not symmetric, means
            diff(x, y) != diff(y, x)
      however
            diff(x, y, how='left') == diff(y, x, how='right')

      Ref: /programming/18180763/set-difference-for-pandas/40209800#40209800
    """
    if (df1.columns != df2.columns).any():
        raise ValueError("Two dataframe columns must match")

    if df1.equals(df2):
        return None
    elif how == 'right':
        return pd.concat([df2, df1, df1]).drop_duplicates(keep=False)
    elif how == 'left':
        return pd.concat([df1, df2, df2]).drop_duplicates(keep=False)
    else:
        raise ValueError('how parameter supports only "left" or "right keywords"')

例:

df1 = pd.DataFrame(d1)
Out[1]: 
                Comment  Name  isEnrolled  score
0  He was late to class  Jack        True   2.17
1             Graduated  Nick       False   1.11
2                         Zoe        True   4.12


df2 = pd.DataFrame(d2)

Out[2]: 
                Comment  Name  isEnrolled  score
0  He was late to class  Jack        True   2.17
1           On vacation   Zoe        True   4.12

diff_df(df1, df2)
Out[3]: 
     Comment  Name  isEnrolled  score
1  Graduated  Nick       False   1.11
2              Zoe        True   4.12

diff_df(df2, df1)
Out[4]: 
       Comment Name  isEnrolled  score
1  On vacation  Zoe        True   4.12

# This gives the same result as above
diff_df(df1, df2, how='right')
Out[22]: 
       Comment Name  isEnrolled  score
1  On vacation  Zoe        True   4.12

A function that finds asymmetrical difference between two data frames is implemented below: (Based on set difference for pandas) GIST: https://gist.github.com/oneryalcin/68cf25f536a25e65f0b3c84f9c118e03

def diff_df(df1, df2, how="left"):
    """
      Find Difference of rows for given two dataframes
      this function is not symmetric, means
            diff(x, y) != diff(y, x)
      however
            diff(x, y, how='left') == diff(y, x, how='right')

      Ref: https://stackoverflow.com/questions/18180763/set-difference-for-pandas/40209800#40209800
    """
    if (df1.columns != df2.columns).any():
        raise ValueError("Two dataframe columns must match")

    if df1.equals(df2):
        return None
    elif how == 'right':
        return pd.concat([df2, df1, df1]).drop_duplicates(keep=False)
    elif how == 'left':
        return pd.concat([df1, df2, df2]).drop_duplicates(keep=False)
    else:
        raise ValueError('how parameter supports only "left" or "right keywords"')

Example:

df1 = pd.DataFrame(d1)
Out[1]: 
                Comment  Name  isEnrolled  score
0  He was late to class  Jack        True   2.17
1             Graduated  Nick       False   1.11
2                         Zoe        True   4.12


df2 = pd.DataFrame(d2)

Out[2]: 
                Comment  Name  isEnrolled  score
0  He was late to class  Jack        True   2.17
1           On vacation   Zoe        True   4.12

diff_df(df1, df2)
Out[3]: 
     Comment  Name  isEnrolled  score
1  Graduated  Nick       False   1.11
2              Zoe        True   4.12

diff_df(df2, df1)
Out[4]: 
       Comment Name  isEnrolled  score
1  On vacation  Zoe        True   4.12

# This gives the same result as above
diff_df(df1, df2, how='right')
Out[22]: 
       Comment Name  isEnrolled  score
1  On vacation  Zoe        True   4.12

回答 12

将pda导入为pd将numpy导入为np

df = pd.read_excel(’D:\ HARISH \ DATA SCIENCE \ 1 MY Training \ SAMPLE DATA&PROJS \ CRICKET DATA \ IPL PLAYER LIST \ IPL PLAYER LIST _ harish.xlsx’)

df1 = srh = df [df [‘TEAM’]。str.contains(“ SRH”)] df2 = csk = df [df [‘TEAM’]。str.contains(“ CSK”)]

srh = srh.iloc [:,0:2] csk = csk.iloc [:,0:2]

csk = csk.reset_index(drop = True)csk

srh = srh.reset_index(drop = True)srh

new = pd.concat([srh,csk],axis = 1)

new.head()

**玩家类型玩家类型

0戴维·华纳·蝙蝠侠… MS Dhoni Captain

1布瓦内什瓦尔·库马尔·鲍勒(Bhuvaneshwar Kumar Bowler)…

2 Manish Pandey击球手… Suresh Raina All-Rounder

3拉希德·汗·阿曼·鲍勒(Kashir Jadhav All-Rounder)

4 Shikhar Dhawan击球手…. Dwayne Bravo All-Rounder

import pandas as pd
import numpy as np

df = pd.read_excel('D:\\HARISH\\DATA SCIENCE\\1 MY Training\\SAMPLE DATA & projs\\CRICKET DATA\\IPL PLAYER LIST\\IPL PLAYER LIST _ harish.xlsx')


df1= srh = df[df['TEAM'].str.contains("SRH")]
df2 = csk = df[df['TEAM'].str.contains("CSK")]   

srh = srh.iloc[:,0:2]
csk = csk.iloc[:,0:2]

csk = csk.reset_index(drop=True)
csk

srh = srh.reset_index(drop=True)
srh

new = pd.concat([srh, csk], axis=1)

new.head()
** 
               PLAYER     TYPE           PLAYER         TYPE

0        David Warner  Batsman    ...        MS Dhoni      Captain

1  Bhuvaneshwar Kumar   Bowler  ...    Ravindra Jadeja  All-Rounder

2       Manish Pandey  Batsman   ...   Suresh Raina  All-Rounder

3   Rashid Khan Arman   Bowler     ...   Kedar Jadhav  All-Rounder

4      Shikhar Dhawan  Batsman    ....    Dwayne Bravo  All-Rounder