Question

我有两个数据帧。例子：

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

每个数据框都以日期作为索引。两个数据帧都具有相同的结构。

我想要做的是比较这两个数据帧并找出df2中哪些行不在df1中。我想比较日期（索引）和第一列（Banana，APple等），看看它们是否存在于df2 vs df1中。

我尝试了以下内容：

对于第一种方法，我收到此错误： “异常：只能比较标记相同的DataFrame对象” 。我已经尝试删除日期作为索引，但得到相同的错误。

在third approach上，我得到断言返回False，但无法弄清楚如何实际看到不同的行。

欢迎提出任何指示

Answer 1

此方法df1 != df2仅适用于具有相同行和列的数据框。实际上，所有数据帧轴都与_indexed_same方法进行比较，如果发现差异，则会引发异常，即使在列/索引顺序中也是如此。

如果我找到了你，你想要找不到变化，而是对称差异。为此，一种方法可能是连接数据帧：

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

分组

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

获取唯一记录的索引

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

过滤

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

Answer 2

将数据帧传递到字典中的concat，从而生成一个多索引数据框，您可以从中轻松删除重复项，从而产生具有数据帧之间差异的多索引数据框：

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)

结果：

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

Answer 3

将 ling 对上面 jur 的回复的评论更新和放置到其他人更容易找到的地方。

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

使用这些数据框进行测试：

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'],
})

结果是：

Answer 4

基于alko的答案几乎对我有用，除了过滤步骤（我得到的地方：ValueError: cannot reindex from a duplicate axis），这是我使用的最终解决方案：

# 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]

Answer 5

有一个更简单的解决方案，更快更好，如果数字不同甚至可以给你数量差异：

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_'])

这里df_diff是差异的概要。您甚至可以使用它来查找数量的差异。在您的示例中：

说明：与比较两个列表类似，为了有效地做到这一点，我们应该首先对它们进行排序然后比较它们（将列表转换为集合/散列也会很快;两者都是对简单的O（N ^ 2）双重比较循环的一个令人难以置信的改进

注意：以下代码生成表：

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'],
})

Answer 6

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

Answer 7

自pandas >= 1.1.0起，我们拥有DataFrame.compare和Series.compare。

注意：该方法只能比较标记相同的DataFrame对象，这意味着具有相同行和列标签的数据框。

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

Answer 8

我得到了这个解决方案。这对你有帮助吗？

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-Z\d]+).*\n'
               '(20\d\d-[01]\d-[0123]\d.+\n?'
               '(.+\n?)*)'
               '(?=[ \n]*\Z'
                  '|'
                  '\n+[a-zA-Z\d]+.*\n'
                  '20\d\d-[01]\d-[0123]\d)')

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

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\n%s\n' % (name,'\n'.join(lu))

结果

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

Answer 9

需要注意的一个重要细节是，您的数据具有重复的索引值，因此要执行任何直接比较，我们需要将所有内容与df.reset_index()一起变为唯一，因此我们可以执行基于选择的选择在条件上。一旦你的情况下定义了索引，我假设你想保留de index所以有一个单行解决方案：

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

一旦从pythonic的角度来看，目标是提高可读性，我们可以稍微打破一下：

# 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)

Answer 10

在这里创建一个简单的解决方案：

Factforge overview

LIMIT

Answer 11

希望这对您有用。 ^ 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):\n{df1}\ndf2:\n{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):\n{df1}")
"""
df1(After):
   date  col1
0  0207     1
1  0207     2
2  0208     3
3  0208     4
"""

Answer 12

我尝试了这种方法，并且有效。我希望它也能提供帮助：

"""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']]

Answer 13

# 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']

Answer 14

将现有数据从 df2 获取到 df1：

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

从df2获取不存在的数据到df1：

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

您可以使用多个比较。

Answer 15

您可以找到 DataFrame 行数之间的差异：

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

输出：

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

比较两个数据帧并获得差异

15 个答案: