下面是另一种使用选择和合并的方法:

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

以下是来自Jupyter的截图:

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)]。

这个答案只是扩展了@Andy Hayden的答案，使其能够适应数值字段为nan的情况，并将其包装成一个函数。

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)

所以对于你的数据(稍微编辑一下，在分数列中有一个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

下面是另一种使用选择和合并的方法:

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

以下是来自Jupyter的截图:

在摆弄了@journois的答案后，由于Panel的弃用，我能够使用MultiIndex而不是Panel来工作。

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，定义你的diff函数，在这种情况下，我将使用他的答案report_diff保持不变:

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

然后，我将数据连接到一个MultiIndex数据框架:

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，我将对每个列组应用report_diff:

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

这个输出:

         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

这就是全部!

第一部分类似于Constantine，你可以得到哪个行是空的布尔值*:

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

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

然后我们可以看到哪些条目发生了变化:

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

这里的第一个条目是索引，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

*注意:重要的是df1和df2在这里共享相同的索引。为了克服这种模糊性，可以使用df1确保只查看共享标签。Index & df2。索引，但我还是把它留作练习吧。