如何使用python比较两列并计算数据帧中有多少相同的项/字符串？

如何使用python比较两列并计算数据帧中有多少相同的项/字符串？ 例如：

row | column A | column B |
============================
 1  | ['NNP',  | ['NNP',  |
    |  'NNP',  | 'NNP',   |
    |  'NNP',  | 'VB',    |
    |  'NNP',  | 'NN',    |
    |  'CC',   | 'NN',    |
    |  'RB',   | 'Z']     |
    |  'NN',   |          |
    |  'Z',    |          |
2   | ['NNP',  | ['NNP',  |
    |  'VB',   |  'NN',   |
    |  'NN',   |  'VB']   |
    |  'NN',   |          |
    |  'Z']    |          |

我想得到的是：

   row | column A | column B |  count_same_string
    ==============================================
     1  | ['NNP',  | ['NNP',  | 4
        |  'NNP',  | 'NNP',   |
        |  'NNP',  | 'VB',    |
        |  'NNP',  | 'NN',    |
        |  'CC',   | 'NN',    |
        |  'RB',   | 'Z']     |
        |  'NN',   |          |
        |  'Z',    |          |
    2   | ['NNP',  | ['NNP',  |2
        |  'VB',   |  'NN',   |
        |  'NN',   |  'RB']   |
        |  'NN',   |          |
        |  'Z']    |          |

---

您可以使用以下代码实现这一点

df['count_same_string'] = df.apply(lambda row: len(set(row['column A']).intersection(row['column B'])))

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您可以使用列表理解修改答案并检索长度：

from collections import Counter

df = pd.DataFrame({"column A":[["NNP","NNP","NNP","NNP","CC","RB","NN","Z"]],
                   "column B":[["NNP","NNP","VB","NN","NN","Z"]]})

df["result"] = [len(list((Counter(a) & Counter(b)).elements()))
                for a,b in zip(df["column A"], df["column B"])]

print (df)

                              column A                   column B  result
0  [NNP, NNP, NNP, NNP, CC, RB, NN, Z]  [NNP, NNP, VB, NN, NN, Z]       4

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你是如何得到第一行的答案的

VB

和

RB

不应该是相同的？？