python - 将两列 Pandas 列表彼此分开

Question

我有一个这样的 df:

col1        col2
[1,3,4,5]   [3,3,6,2]
[1,4,5,5]   [3,8,4,3]
[1,3,4,8]   [8,3,7,2]

尝试将 col1 和 col2 中的列表中的元素划分在一起以获得结果列中的内容:

col1        col2        result
[1,3,4,5]   [3,3,6,2]   [.33,1,.66,2.5]
[1,4,5,5]   [3,8,4,3]   [.33,.5,1.25,1.66]
[1,3,4,8]   [8,3,7,2]   [.33,1,.57,4]

尝试了很多不同的方法 - 但总是出错。
尝试 :

#attempt1
df['col1'].div(df['col2'], axis=0)

#attempt2
from operator import truediv

for i in df.col1:
     a = np.array(df['col1'])
     for t in df.col2:
         b = np.array(df['col2'])
         x = a/b
         print(x)


#attempt3
for i in df.index:
    a = col1
    b = col2
    x = map(truediv, a, b)

#attempt4
a = col1
b = col2
result = [x/y for x, y in zip(a, b)]
#then apply to df

#attempt5
a = col1
b = col2
result = a/b
print(percent_matched)
#then #apply to df

>>>TypeError: unsupported operand type(s) for /: 'list' and 'list'

有任何想法吗？

Answer 1

您可以将列表理解与应用一起使用，这取决于两个列表的长度相同

df['result'] = df.apply(lambda x: [np.round(x['col1'][i]/x['col2'][i], 2) for i in range(len(x['col1']))], axis = 1)

    col1            col2            result
0   [1, 3, 4, 5]    [3, 3, 6, 2]    [0.33, 1.0, 0.67, 2.5]
1   [1, 4, 5, 5]    [3, 8, 4, 3]    [0.33, 0.5, 1.25, 1.67]
2   [1, 3, 4, 8]    [8, 3, 7, 2]    [0.12, 1.0, 0.57, 4.0]

编辑:正如@TrentonMcKinney 所建议的，这可以在不使用 LC 的情况下完成。该解决方案利用了 Numpy 的矢量化操作，

df.apply(lambda x: np.round(np.array(x[0]) / np.array(x[1]), 3), axis=1)