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)

打印

                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

我遇到过这个问题，但在找到这篇文章之前找到了答案:

根据unutbu的回答，加载你的数据…

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])

定义你的diff函数…

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

然后你可以简单地使用Panel来总结:

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

顺便说一下，如果你在IPython Notebook中，你可能喜欢使用彩色差异函数 根据单元格是否不同、相等或左/右为空来给出颜色:

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))

pandas >= 1.1: DataFrame.compare

使用pandas 1.1，基本上可以用一个函数调用复制Ted Petrou的输出。例子摘自文档:

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

这里，“self”指的是LHS数据帧，而“other”指的是RHS数据帧。默认情况下，相等的值将被nan替换，因此您可以只关注差异。如果您想显示相同的值，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

你也可以使用align_axis改变比较轴:

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

这是逐行比较值，而不是逐列比较值。

如果两个数据帧中有相同的id，那么找出发生了什么变化实际上是相当容易的。只要执行frame1 != frame2，就会得到一个布尔型的DataFrame，其中每个True都是已更改的数据。由此，您可以通过执行changedids = frame1.index[np。Any (frame1 != frame2,axis=1)]。

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()

** 玩家类型 0 David Warner Batsman…多尼女士，机长 1 Bhuvaneshwar Kumar Bowler…拉文德拉·加德贾是全才 Manish Pandey Batsman…苏雷什·莱纳全能 拉希德·汗·阿尔曼·鲍勒…基达尔·贾达夫全能 4 Shikhar Dhawan Batsman ....多面手Dwayne Bravo

扩展@cge的答案，这对于结果的可读性来说非常酷:

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('')

结果:样本

在线演示