Question

如何获取列表中每个字符串的第一个单词并创建一个新列表。

list1 = ['GenX XE','GenX XM','GenX XT','GenX XMA','GenX XTA']

# this is my my list 
n= len(list1)

for i in range(n):
    print (list1[i])

输出：

[GenX', 'GenX', 'GenX', 'GenX', 'GenX']

Answer 1

使用list_comprehensions：

In [1757]: n = [i.split()[0] for i in list1]

In [1758]: n
Out[1758]: ['GenX', 'GenX', 'GenX', 'GenX', 'GenX']

这基本上是指：

In [1754]: n = []

In [1755]: for i in list1:
      ...:     n.append(i.split()[0])
      ...:     

In [1756]: n
Out[1756]: ['GenX', 'GenX', 'GenX', 'GenX', 'GenX']

Answer 2

使用str.split，然后使用索引访问元素。在下面的示例中，我使用列表理解

例如：

list1 = ['GenX XE','GenX XM','GenX XT','GenX XMA','GenX XTA']
print([i.split()[0] for i in list1])

输出：

['GenX', 'GenX', 'GenX', 'GenX', 'GenX']

Answer 3

您可以这样做

In [5]: map(lambda x:x.split()[0],list1)
Out[5]: ['GenX', 'GenX', 'GenX', 'GenX', 'GenX']

Answer 4

list1 = ['GenX XE','GenX XM','GenX XT','GenX XMA','GenX XTA']

# this is my my list 
n= len(list1)

for i in range(n):
     print (list1[i].split()[0])

注意：list1 [i] .split（）[0]

Answer 5

您可以使用正则表达式执行此操作：

import re
import numpy as np
l = ['GenX XE','GenX XM','GenX XT','GenX XMA','GenX XTA']
r = [re.findall('(?i)([A-Z0-9]+)',x) for x in l]

要提取特定信息，您可以执行以下操作：

np.array(r)[:,0]

并获取：

array(['GenX', 'GenX', 'GenX', 'GenX', 'GenX'], dtype='<U4')

如何删除列表中每个字符串的第一个单词并将这些单词附加到新列表中

5 个答案: