如何检查 df1 行与另一个 df2 列的所有单行以检查数据框中是否相同

Question

我见过很多这样的代码：

mergedStuff = pd.merge(df1, df2, how='inner')

or

mask = df1.reindex(df2.index).values == df2.values

但以上仅在同一行存在于其相对行中时才给出它不与每一行进行比较

例如：

 df1 contains:
  
  0

  hello  
  how
  are
  you 
  guys
  system

  df2 contains:
  
  0             1   ........ n

  how           hello        you
  hello         guys         hello
  you           system       how 
  are           you          you
  guys          hello        hello
  system        how          how
  hello         are          system

更新：比较 df2.isin(df1) 后的最终输出列：

  #NOTE ! below output are manually entered, not from real output
  #        but i know this is how it gives.
  false   True   false
  false   false  false
  false   false  false
  false   false  false
  false   false  false
  false   false  false
  false   false  false      #only second column and second row are 
                                      #  true, because 
                                       # it matches 
                                       # the same row

但我想要的是交叉检查每个 df1 行和 df2 行。

预期输出：

  True  True  True
  True  True  True
  True  True  True
  True  True  True
  True  True  True
  True  True  True
  True  True  True    #i want true for all Because every rows has 
                       # the same word.

更新2：

但是，如果我这样运行，那么它会给出预期的输出：

df2[2].isin(df1[0])

 True
 True
 True
 True
 True
 True
 True  # 2nd column of df2 compared with df1 and gives good output.
        # but if i give without index it gives crap.

如果你想帮忙，这里是测试的输入：

df1 = pd.DataFrame({0: 
['hello','how','are','you','guys','system']})

   df2 = pd.DataFrame({ 0: 
  ['how','hello','you','hello','guys','hello',
    'you','system','how','are','you','you' 
     'guys','hello','hello','system','how',
      'how','hello','are','system'],

    1: ['how','you','you','hello','guys',
      'hello','you','system','how','are','you','you' 
       'guys','hello','hello','system',
       'how','hello','hello','are','system'] ,

      2: ['how','you','you','are','guys',
     'hello','you','system','you','are','guys','you' 
          'guys','hello','hello','system',
        'how','hello','hello','are','system']
            })

这个终于成功了：

new = np.isin(df2, df1)
rows, cols = np.nonzero(~new)

#or

new = np.isin(df2, df1, invert=True)
rows, cols = np.nonzero(new)


x2 = []

for item in zip(rows,cols):
    x2.append(df2.iloc[item]) 

Answer 1

试试pandas.DataFrame.isin

df1 = pd.DataFrame(['hello', 'how', 'are', 'you'], columns=['column1'])

df2 = pd.DataFrame(['hello', 'are', 'you', 'how'], columns=['column1'])

df1["column1"].isin(df2["column1"])

# 0    True
# 1    True
# 2    True
# 3    True
# Name: column1, dtype: bool

注意：恐怕是O(n*m) 复杂性，这意味着对于 df1 的每个元素，您都在迭代 df2，尽管我不知道实际的实现。更快的方法是对两列进行排序并检查是否相等。

df1 = pd.DataFrame(['hello', 'how', 'are', 'you'], columns=['column1'])

df2 = pd.DataFrame(['hello', 'are', 'you', 'how'], columns=['column1'])

df1.sort_values(['column1'], inplace=True, ignore_index=True)
df2.sort_values(['column1'], inplace=True, ignore_index=True)

df1['column1'] == df2['column1']

0    True
1    True
2    True
3    True
Name: column1, dtype: bool

Answer 2

使用np.in1d:

>>> df2.apply(lambda x: np.in1d(x, df1[0]))
        0      1      2
0    True   True   True
1    True   True   True
2    True   True   True
3    True   True   True
4    True   True   True
5    True   True   True
6    True   True   True
7    True   True   True
8    True   True   True
9    True   True   True
10   True   True   True
11  False  False  False
12   True   True   True
13   True   True   True
14   True   True   True
15   True   True   True
16   True   True   True
17   True   True   True
18   True   True   True
19   True   True   True

更新：缺失单词列表

>>> set(np.extract(mask == False, df2))
{'youguys'}

Answer 3

new = np.isin(df2, df1)
rows, cols = np.nonzero(~new)

#or

new = np.isin(df2, df1, invert=True)
rows, cols = np.nonzero(new)


x2 = []

for item in zip(rows,cols):
    x2.append(df2.iloc[item])