blocks|key|1703384|text|如果regex不是必需的，您可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1703385|l+=+text.split('+')
out+=+[]
for+i+in+range(len(l)):
++++try:
++++++++o.append(l[i]+%2B+'+'+%2B+l[i%2B1])
++++except+IndexError:
++++++++continue|code-block|syntax|javascript|1703386|解释：|1703387|首先拆分空格字符上的字符串。结果将是一个列表，其中每个元素都是句子中的一个单词。实例化一个空列表以保存结果。循环遍历单词列表，将由空格分隔的两个单词组合添加到输出列表中。这将在访问列表中的最后一个单词时抛出一个IndexError，只需捕获它并继续，因为您似乎不希望结果中只有这个单词。|1703388|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

If regex isn't require you could do something like:

<pre class="lang-py prettyprint-override"><code>l = text.split(' ')
out = []
for i in range(len(l)):
 try:
 o.append(l[i] + ' ' + l[i+1])
 except IndexError:
 continue
</code></pre>

Explanation:

First split the string on the space character. The result will be a list where each element is a word in the sentence. Instantiate an empty list to hold the result. Loop over the list of words adding the two word combinations seperated by a space to the output list. This will throw an IndexError when accessing the last word in the list, just catch it and continue since you don't seem to want that lone word in your result anyway.

blocks|key|1949904|text|我不认为你真的需要这个regex。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1949905|我知道您需要一个列表，其中每个元素包含两个单词，后者也是以下元素中的前一个。我们可以很容易地做到这样：|1949906|string+=+"Hello+there+how+are+you"
liste+=+string.split("+").pop(-1)
#+we+remove+the+last+index,+as+otherwise+we'll+crash,+or+have+an+element+with+only+one+word
for+i+in+range(len(liste)-1):
++++liste[i]+=+liste[i]+%2B+"+"+%2B+liste[i%2B1]|code-block|syntax|javascript|1949907|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I don't think you actually need regex for this. 
I understand you want a list, in which each element contains two words, the latter also being the former of the following element. We can do this easily like this: 

<pre class="lang-py prettyprint-override"><code>string = "Hello there how are you"
liste = string.split(" ").pop(-1)
# we remove the last index, as otherwise we'll crash, or have an element with only one word
for i in range(len(liste)-1):
 liste[i] = liste[i] + " " + liste[i+1]
</code></pre>

blocks|key|1949921|text|我不知道你是否必须使用regex，但我会这样做。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1949922|首先，您可以使用str.split()方法获取单词列表。|offset|length|style|CODE|1949923|>>>+sentence+=+"hello+there+how+are+you"
>>>+splited_sentence+=+sentence.split("+")
>>>+splited_sentence
['hello',+'there',+'how',+'are',+'you']|code-block|syntax|javascript|1949924|然后，你可以做成对。|1949925|>>>+output+=+[]
>>>+for+i+in+range+(1,+len(splited_sentence)+):
...+++++output+%2B=+[+splited[+i-1+]+%2B+'+'+%2B+splited_sentence[+i+]+]
...
output
['hello+there',+'there+how',+'how+are',+'are+you']|1949926|entityMap^0|0|8|B|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|W|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|Y|8|@]|9|@]|A|$K|L]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

I don't know if it's mandatory for you need to use regex, but I'd do this way.

First, you can get the list of words with the <code>str.split()</code> method.

<pre class="lang-py prettyprint-override"><code>&gt;&gt;&gt; sentence = "hello there how are you"
&gt;&gt;&gt; splited_sentence = sentence.split(" ")
&gt;&gt;&gt; splited_sentence
['hello', 'there', 'how', 'are', 'you']
</code></pre>

Then, you can make pairs.

<pre class="lang-py prettyprint-override"><code>&gt;&gt;&gt; output = []
&gt;&gt;&gt; for i in range (1, len(splited_sentence) ):
... output += [ splited[ i-1 ] + ' ' + splited_sentence[ i ] ]
...
output
['hello there', 'there how', 'how are', 'are you']
</code></pre>

blocks|key|1949976|text|下面是使用re.findall的解决方案|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1949977|>>>+import+re
>>>+text+=+"hello+there+how+are+you"
>>>+re.findall(r"(?=(?:(?:%5E%7C\W)(\S%2B\W\S%2B)(?:$%7C\W)))",+text)
['hello+there',+'there+how',+'how+are',+'are+you']|code-block|syntax|javascript|1949978|查看re：https://docs.python.org/3/library/re.html的Python|1949979|(?=...)前瞻性断言|unordered-list-item|1949980|(?:...)非捕获规则括号|1949981|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/re.html^0|5|A|0|0|2|2|5|15|0|0|0|7|0|0|7|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@$9|16|A|17|1|18]]|E|$]]|$1|M|3|N|5|O|7|19|8|@$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|P|3|Q|5|O|7|1C|8|@$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|R|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|S|$T|$5|U|V|W|E|$X|Y]]]]

Here's a solution with <code>re.findall</code>:

<pre class="lang-py prettyprint-override"><code>&gt;&gt;&gt; import re
&gt;&gt;&gt; text = "hello there how are you"
&gt;&gt;&gt; re.findall(r"(?=(?:(?:^|\W)(\S+\W\S+)(?:$|\W)))", text)
['hello there', 'there how', 'how are', 'are you']
</code></pre>

Have a look at the Python docs for <code>re</code>: <a href="https://docs.python.org/3/library/re.html" rel="nofollow noreferrer">https://docs.python.org/3/library/re.html</a>

<ul>
<li><code>(?=...)</code> Lookahead assertion</li>
<li><code>(?:...)</code> Non-capturing regular parentheses</li>
</ul>

blocks|key|4222643|text|使用findall的另一个可能的解决方案。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4222644|>>>+liste+=+list(map(''.join,+re.findall(r'(\S%2B(?=(\s%2B\S%2B)))',+text)))
>>>+liste
['hello+there',+'there+how',+'how+are',+'are+you']|code-block|syntax|javascript|4222645|entityMap^0|2|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Another possible solution using <code>findall</code>.

<pre><code>&gt;&gt;&gt; liste = list(map(''.join, re.findall(r'(\S+(?=(\s+\S+)))', text)))
&gt;&gt;&gt; liste
['hello there', 'there how', 'how are', 'are you']
</code></pre>

blocks|key|1950046|text|另一种选择就是split，zip，join，就像.|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1950047|sentence+=+"Hello+there+how+are+you"
words+=+sentence.split()
['+'.join(i)+for+i+in+zip(words,+words[1:])]|code-block|syntax|javascript|1950048|entityMap^0|7|5|D|3|H|4|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

An alternative is just to <code>split</code>, <code>zip</code>, then <code>join</code> like so...

<pre><code>sentence = "Hello there how are you"
words = sentence.split()
[' '.join(i) for i in zip(words, words[1:])]
</code></pre>

What is the best way to split a string like

<pre><code>text = "hello there how are you"
</code></pre>

in Python?

So I'd end up with an array like such:

<pre><code>['hello there', 'there how', 'how are', 'are you']
</code></pre>

I have tried this:

<pre><code>liste = re.findall('((\S+\W*){'+str(2)+'})', text)
for a in liste:
 print(a[0])
</code></pre>

But I'm getting:

<pre><code>hello there 
how are 
you
</code></pre>

How can I make the <code>findall</code> function move only one token when searching?

split string by using regex in python

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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拆分字符串的最佳方法是什么？text = "hello there how are you"用Python？所以我最终会得到这样的数组：['hello there', 'there how', 'how are', 'are you']我试过这样做：liste = re.findall('((\S+\W*){'+str...

问在python中使用regex拆分字符串
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