获取列表中所有字符串的第一个单词

Question

我有一个正在阅读的 CSV 文件，如下所示。我需要得到所有字符串的第一个字。我知道如何获得首字母，但我不确定如何获得单词。

['diffuse systemic sclerosis', 'back', 'public on july 15 2008']
['diffuse systemic sclerosis', 'forearm', 'public on may 9 2014']

我希望我的输出是

diffuse
back
public
forearm

Answer 1

您可以使用列表理解和 split() 函数：

>>> l=['diffuse systemic sclerosis', 'back', 'public on july 15 2008']
>>> [i.split()[0] for i in l]
['diffuse', 'back', 'public']

Answer 2

你可以使用理解

>>> l = [['diffuse systemic sclerosis', 'back', 'public on july 15 2008']
,['diffuse systemic sclerosis', 'forearm', 'public on may 9 2014']]

>>> list({i.split()[0] for j in l for i in j})
['back', 'diffuse', 'forearm', 'public']

Answer 3

l = [
    ['diffuse systemic sclerosis', 'back', 'public on july 15 2008'],
    ['diffuse systemic sclerosis', 'forearm', 'public on may 9 2014']
    ]
d = lambda o: [a.split().pop(0) for a in o]
r = lambda a,b: d(a) + d(b)
print "\n".join(set(reduce(r, l)))
>>> 
public
forearm
diffuse
back

Answer 4

可以在列表理解中使用str.split，注意可以指定maxsplit来减少操作次数：

L = ['diffuse systemic sclerosis', 'back', 'public on july 15 2008']

res = [i.split(maxsplit=1)[0] for i in L]
# ['diffuse', 'back', 'public']

您也可以在功能上执行相同的操作：

from operator import itemgetter, methodcaller

splitter = methodcaller('split', maxsplit=1)
res = list(map(itemgetter(0), map(splitter, L)))

在多个列表中，如果您希望保持观察第一个单词的顺序，您可以使用 itertool unique_everseen recipe, also found in the more_itertools 库：

from itertools import chain
from more_itertool import unique_everseen

L1 = ['diffuse systemic sclerosis', 'back', 'public on july 15 2008']
L2 = ['diffuse systemic sclerosis', 'forearm', 'public on may 9 2014']

res = list(unique_everseen(i.split(maxsplit=1)[0] for i in chain(L1, L2)))

# ['diffuse', 'back', 'public', 'forearm']

获取列表中所有字符串的第一个单词

Get first word of all strings in lists

python

string

split

list

strip