python - 将列表的两个熊猫列相互除以

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？

标签： pythonpandas

Answer 1

• 用于.applymap将列转换为np.arrays
• 然后用于.div划分列
• 如果result必须四舍五入，则.apply(lambda x: np.round(x, 3))在计算该列时添加 。
  • np.round()
  • df['result'] = df.col1.div(df.col2).apply(lambda x: np.round(x, 3))

import numpy as np
import pandas as pd

data = {'col1': [[1,3,4,5], [1,4,5,5], [1,3,4,8]], 'col2': [[3,3,6,2], [3,8,4,3], [8,3,7,2]]}

df = pd.DataFrame(data)

# convert columns to arrays
df = df.applymap(np.array)

# divide the columns
df['result'] = df.col1.div(df.col2)