Python 如何将列表中的元素合并在一起？

我有这个句子列表，我想把它们合并在一起，创建一个完整的列表

test = [['Hello my name is Py. How are you today?'],['The world is a great place. Another sentence.']]

我如何合并元素来创建它

test = ['Hello my name is Py. How are you today? The world is a great place. Another sentence.']

或

谢谢，您可以使用来链接每个子列表中的元素，在链对象上调用str.join来创建单个字符串

test = [['Hello my name is Py. How are you today?'],['The world is a great place. Another sentence.']]

from itertools import chain

print(" ".join(chain.from_iterable(test)))
Hello my name is Py. How are you today? The world is a great place. Another sentence

或者仅使用“加入”：

print(" ".join(["".join(sub) for sub in test]))
Hello my name is Py. How are you today? The world is a great place. Another sentence.

如果每个子列表中只有一个sting，则只需索引：

print(" ".join([sub[0] for sub in test]))

Hello my name is Py. How are you today? The world is a great place. Another sentence.

要获取列表，只需将联接包装到列表中：

print([" ".join([sub[0] for sub in test])])
['Hello my name is Py. How are you today? The world is a great place. Another sentence.']

如果每个子列表中都有很多子字符串，则链将是最有效的解决方案。

要获取列表-

>>> test = ["".join([j  for j in i for i in test])]
>>> test
['The world is a great place. Another sentence.The world is a great place. Another sentence.']

串

>>> test = "".join([j  for j in i for i in test])
>>> test
'The world is a great place. Another sentence.The world is a great place. Another sentence.'

---

连接列表，然后连接字符串：

' '.join(sum(test, []))

演示：

警告：虽然这对于一些列表来说很简单，但列表越多，速度越慢：

>>> for n in (10, 100, 1000, 10000, 100000):
        lists = [['Test'] for _ in range(n)]
        seconds = timeit(lambda: sum(lists, []), number=1)
        print('%10.7f' % seconds, 'seconds for', n, 'lists')


 0.0000109 seconds for 10 lists
 0.0001052 seconds for 100 lists
 0.0053068 seconds for 1000 lists
 0.5582595 seconds for 10000 lists
55.8725820 seconds for 100000 lists

普通列表理解速度更快：

>>> for n in (10, 100, 1000, 10000, 100000):
        lists = [['Test'] for _ in range(n)]
        seconds = timeit(lambda: [e for s in lists for e in s], number=1)
        print('%10.7f' % seconds, 'seconds for', n, 'lists')

 0.0000115 seconds for 10 lists
 0.0000327 seconds for 100 lists
 0.0002784 seconds for 1000 lists
 0.0024991 seconds for 10000 lists
 0.0228550 seconds for 100000 lists

---

这将实现以下目的：

test = [['Hello my name is Py. How are you today?'],['The world is a great place. Another sentence.']]
import itertools
print list(itertools.chain(*test))

---

为什么每个字符串都有自己的子列表？@marmeladze，是的，对于多个子字符串来说，这将是最有效的。你的理解是反向的。我认为最好更精确一些，说“汇总列表”而不是“加入列表”，因为

join（）

只处理字符串和

sum（）

拒绝连接字符串。嗯。。。我认为“加入”是一个更自然的词。我不是在谈论Python函数名，而是在使用英语单词。我现在做了一个谷歌测试，“加入两个列表”的结果大约是“两个列表相加”结果的十倍。如果您觉得更好的话，我会切换到“concatenate”，但我真的不想说“sum”，因为这是我使用的Python函数的名称。“Join”确实是一个自然词，但当您使用它来指代

sum（）

，它会掩盖

Join（）

：P由于我过去曾尝试使用

sum（）

和

join（）

作为可互换的工具，因此我发现在心理上区分它们很有帮助。好吧，我不想被称为伪装者，所以我改为“串联”。并添加了一个关于速度的警告/测试。它不会，它将创建一个包含多个子字符串而不是单个字符串的列表

>>> for n in (10, 100, 1000, 10000, 100000):
        lists = [['Test'] for _ in range(n)]
        seconds = timeit(lambda: sum(lists, []), number=1)
        print('%10.7f' % seconds, 'seconds for', n, 'lists')


 0.0000109 seconds for 10 lists
 0.0001052 seconds for 100 lists
 0.0053068 seconds for 1000 lists
 0.5582595 seconds for 10000 lists
55.8725820 seconds for 100000 lists

>>> for n in (10, 100, 1000, 10000, 100000):
        lists = [['Test'] for _ in range(n)]
        seconds = timeit(lambda: [e for s in lists for e in s], number=1)
        print('%10.7f' % seconds, 'seconds for', n, 'lists')

 0.0000115 seconds for 10 lists
 0.0000327 seconds for 100 lists
 0.0002784 seconds for 1000 lists
 0.0024991 seconds for 10000 lists
 0.0228550 seconds for 100000 lists

test = [['Hello my name is Py. How are you today?'],['The world is a great place. Another sentence.']]
import itertools
print list(itertools.chain(*test))