Question

目前，这有效：

df['new'] = df.apply( \
       lambda x: address[int(x['c1'][:5], 2)]+'_'+str(int(x['c1'][6:11], 2)) \
       if x['c1'][5] == '1' \
       else address[int(x['c2'][:5], 2)]+'_'+str(int(x['c2'][6:11], 2)), axis=1) `

address是一本字典。

但非常慢。具体而言，apply对整个数据帧的速度比apply到选定列慢得多。但是，新列基于多列，我不知道如何实现它。

此外，有没有办法对这些类型的逻辑/条件语句进行矢量化？

示例数据框： <bound method DataFrame.head of c1 c2 0 0000100111000111 0010110011000111 1 0001000111000111 0010110011000111 2 0101010001001010 0000000000000000 3 0101010010001110 0000000000000000 4 0101010011101010 0000000000000000 5 0111111100000100 0000000000000000 6 0111110010010110 0000000000000000 7 1000000001001100 0000000000000000 8 1110011110001000 0000000000000000 9 0000100001010000 0000000000000000 10 0001000001001010 0000000000000000 11 0101101100100100 0000000000000000 12 1110001100100100 0000000000000000 13 0010100101101001 0101010101101001 14 0000100101100000 0000000000000000 15 0000100110100000 0000000000000000 16 0001000101101011 0000000000000000 17 1001110000100001 0000000000000000 18 0111111000100000 0000000000000000 19 1000000100010110 0000000000000000 20 1110001111000010 0000000000000000 21 1011010001000010 0000000000000000 22 0110010001001111 0000000000000000 23 0111110000110101 0000000000000000 24 0111110001001100 0000000000000000 25 1000000000111101 0000000000000000 26 0000110001100010 0000000000000000 27 0001010001100010 0000000000000000 28 1100100100100101 1001011000000101 29 0101000010101010 0111110001001010 ... ... ... 95714 0101111100011000 0000000000000000 95715 0010101011001011 0000000000000000 95716 0010100111100110 0101010110100110 95717 0010101000100100 0101011011100100 95718 0101000110000101 0000000000000000

Answer 1

您需要向量化if-then-else，也称为np.where（np代表numpy，以防万一。）

import numpy as np
df['new'] = np.where(df['c1'].str[5] == '1',
                     df['c1'].str[:5], 
                     df['c2'].str[:5])
#                 c1                c2    new
#0  0000100111000111  0010110011000111  00101
#1  0001000111000111  0010110011000111  00101
#2  0101010001001010  0000000000000000  01010
#....

Answer 2

看起来您正在尝试根据字符串列c1的字符值执行操作。像这样进行逐行字符串操作很慢，但是pandas可以帮助你解决.str functions：

# begin by setting all of the values to what you want from c1
df['new'] = df['c1'].str.slice(stop=5)

# replace those that meet your criteria with what you want from 'c2'
df.loc[df['c1'].str.get(5) == '1', 'new'] = df['c2'].str.slice(stop=5)

Answer 3

使用Boolean~

df['New']=df.c1.str[:5]
df.loc[df.c1.str[5]=='1','New']=(df.c2.str[:5])[df.c1.astype(str).str[5]=='1']

优化数据框操作：基于条件逻辑和多列的新列

3 个答案: