在 Python Pandas -> 字符串列表中查找两列的交集
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Find intersection of two columns in Python Pandas -> list of strings
提问by Mia
I would like to count how many instances of column A and B intersect. The rows in Column A and B are lists of strings. For example, column A may contain [car, passenger, truck] and column B may contain [car, house, flower, truck]. Since in this case, 2 strings overlap, column C should display -> 2
我想计算 A 列和 B 列相交的实例数。A 列和 B 列中的行是字符串列表。例如，A 列可能包含 [汽车、乘客、卡车]，B 列可能包含 [汽车、房屋、花卉、卡车]。由于在这种情况下，2 个字符串重叠，因此 C 列应显示 -> 2
I have tried (none of these work):
我试过（这些都不起作用）：
df['unique'] = np.unique(frame[['colA', 'colB']])
or
或者
def unique(colA, colB):
unique1 = list(set(colA) & set(colB))
return unique1
df['unique'] = df.apply(unique, args=(df['colA'], frame['colB']))
TypeError: ('unique() takes 2 positional arguments but 3 were given', 'occurred at index article')
TypeError: ('unique() 需要 2 个位置参数，但给出了 3 个'，'发生在索引文章')
采纳答案by jezrael
I believe need lengthwith set.intersectionin list comprehension:
我相信需要length与set.intersection在列表理解：
df['C'] = [len(set(a).intersection(b)) for a, b in zip(df.A, df.B)]
Or:
或者：
df['C'] = [len(set(a) & set(b)) for a, b in zip(df.A, df.B)]
Sample:
样品：
df = pd.DataFrame(data={'A':[['car', 'passenger', 'truck'], ['car', 'truck']],
'B':[['car', 'house', 'flower', 'truck'], ['car', 'house']]})
print (df)
A B
0 [car, passenger, truck] [car, house, flower, truck]
1 [car, truck] [car, house]
df['C'] = [len(set(a).intersection(b)) for a, b in zip(df.A, df.B)]
print (df)
A B C
0 [car, passenger, truck] [car, house, flower, truck] 2
1 [car, truck] [car, house] 1
