blocks|key|1483021|text|你想要这样的东西：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1483022|l1+=+[['a+b',+'c+d',+'e+f'],+['a+b',+'c+d',+'e+f'],+['a+b',+'c+d',+'e+f']]

l2+=+[]
for+i,j+in+enumerate(l1):
++++l2.append([])
++++for+k+in+j:
++++++++l2[i].extend(k.split())

print(l2)|code-block|syntax|javascript|1483023|DEMO|offset|length|1483024|entityMap|0|LINK|mutability|MUTABLE|url|http://repl.it/kE9^0|0|0|0|4|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

You want something like this:

<pre><code>l1 = [['a b', 'c d', 'e f'], ['a b', 'c d', 'e f'], ['a b', 'c d', 'e f']]

l2 = []
for i,j in enumerate(l1):
 l2.append([])
 for k in j:
 l2[i].extend(k.split())

print(l2)
</code></pre>

<a href="http://repl.it/kE9" rel="nofollow">DEMO</a>

blocks|key|1771677|text|如果您有一个字符串列表|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1771678|tweets+=+['a+tweet',+'another+tweet']|code-block|syntax|javascript|1771679|然后，您可以使用列表理解拆分每个元素|1771680|split_tweets+=+[tweet.split('+')
++++++++++++++++for+tweet+in+tweets]|1771681|因为它是一个tweet列表：|1771682|tweet_groups+=+[['tweet+1',+'tweet+1b'],+['tweet+2',+'tweet+2b']]
tweet_group_words+=+[[word
++++++++++++++++++++++for+tweet+in+group
++++++++++++++++++++++for+word+in+tweet.split('+')]
+++++++++++++++++++++for+group+in+tweet_groups]|1771683|它将给出单词列表的列表。|1771684|如果你想计算不同的单词，|1771685|words+=+[set(word+
+++++++++++++for+tweet+in+group
+++++++++++++for+word+in+tweet.split('+'))
+++++++++for+group+in+tweet_groups]|1771686|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

If you have a list of strings

<pre><code>tweets = ['a tweet', 'another tweet']
</code></pre>

Then you can split each element using a list comprehension

<pre><code>split_tweets = [tweet.split(' ')
 for tweet in tweets]
</code></pre>

Since it's a list of lists of tweets:

<pre><code>tweet_groups = [['tweet 1', 'tweet 1b'], ['tweet 2', 'tweet 2b']]
tweet_group_words = [[word
 for tweet in group
 for word in tweet.split(' ')]
 for group in tweet_groups]
</code></pre>

Which will give a list of lists of words.

If you want to count distinct words,

<pre><code>words = [set(word 
 for tweet in group
 for word in tweet.split(' '))
 for group in tweet_groups]
</code></pre>

blocks|key|1487688|text|groups+=+[["foo+bar",+"bar+baz"],+["foo+foo"]]
[sum((tweet.split('+')+for+tweet+in+group),+[])+for+group+in+groups]
#+=>+[['foo',+'bar',+'bar',+'baz'],+['foo',+'foo']]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1487689|编辑:似乎需要一个解释。|unstyled|1487690|每组[...+for+group+in+groups]的|offset|length|style|CODE|1487691|unordered-list-item|1487692|1487693|-+For+each+tweet,+split+into+words+`(tweet.split('+')+for+tweet+in+group)`
-+Concatenate+the+split+tweets+`sum(...,+[])`|1487694|​|1487695|entityMap^0|0|0|2|P|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|X|8|@$I|Y|J|Z|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|N|7|10|8|@]|9|@]|A|$]]|$1|O|3|-4|5|F|7|11|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|12|8|@]|9|@]|A|$B|C]]|$1|R|3|S|5|F|7|13|8|@]|9|@]|A|$]]|$1|T|3|-4|5|F|7|14|8|@]|9|@]|A|$]]]|U|$]]

<pre><code>groups = [["foo bar", "bar baz"], ["foo foo"]]
[sum((tweet.split(' ') for tweet in group), []) for group in groups]
# =&gt; [['foo', 'bar', 'bar', 'baz'], ['foo', 'foo']]
</code></pre>

EDIT: It seems an explanation is needed.

<ul>
<li>For each group <code>[... for group in groups]</code>

<ul>
<li>For each tweet, split into words <code>(tweet.split(' ') for tweet in group)</code></li>
<li>Concatenate the split tweets <code>sum(..., [])</code></li>
</ul></li>
</ul>

blocks|key|1771722|text|如果你想计算出现的次数，那么使用Counter字典，在拆分之后用itertools.chain链接所有的单词。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1771723|from+collections+import+Counter
from+itertools+import+chain

tweets++=+[['foo+bar',+'foo+foobar'],+['bar+foo',+'bar']]
print([Counter(chain.from_iterable(map(str.split,sub)))++for+sub+in+tweets]+)
[Counter({'foo':+2,+'foobar':+1,+'bar':+1}),+Counter({'bar':+2,+'foo':+1})]|code-block|syntax|javascript|1771724|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/library/collections.html#collections.Counter|1|https://docs.python.org/2/library/itertools.html#itertools.chain^0|G|7|0|W|F|1|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]|$A|W|B|X|1|Y]]|C|$]]|$1|D|3|E|5|F|7|Z|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|10|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]|Q|$5|L|M|N|C|$O|R]]]]

If you want to count the occurrences then use a <a href="https://docs.python.org/2/library/collections.html#collections.Counter" rel="nofollow">Counter</a> dict, chaining all the words with <a href="https://docs.python.org/2/library/itertools.html#itertools.chain" rel="nofollow">itertools.chain</a> after splitting.

<pre><code>from collections import Counter
from itertools import chain

tweets = [['foo bar', 'foo foobar'], ['bar foo', 'bar']]
print([Counter(chain.from_iterable(map(str.split,sub))) for sub in tweets] )
[Counter({'foo': 2, 'foobar': 1, 'bar': 1}), Counter({'bar': 2, 'foo': 1})]
</code></pre>

blocks|key|4089165|text|您可以创建一个函数，将您的列表传递给该函数，该函数将汇编并返回单词的字典，以及它们在您的tweet中出现的次数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4089166|def+countWords(listitem):
++++a+=+[]
++++for+x+in+listitem:
++++++++for+y+in+x:
++++++++++++for+z+in+y.split('+'):
++++++++++++++++a.append(z)
++++b+=+{}
++++for+word+in+a:
++++++++if+word+not+in+b:
++++++++++++b[word]+=+1
++++++++else:
++++++++++++b[word]+%2B=+1
++++return+b|code-block|syntax|javascript|4089167|这样，您将保留您的列表，并能够将返回值分配回一个新变量以供检查。|4089168|dictvar+=+countWords(listoftweets)|4089169|创建定义将允许您将其放入其自己的文件中，以便在将来导入使用。|4089170|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You could create a function that you pass your list to that will assemble and return a dictionary of the words and how many times they show up in your tweets.

<pre><code>def countWords(listitem):
 a = []
 for x in listitem:
 for y in x:
 for z in y.split(' '):
 a.append(z)
 b = {}
 for word in a:
 if word not in b:
 b[word] = 1
 else:
 b[word] += 1
 return b
</code></pre>

this way you will keep both your list and be able to assign the return value back to a new variable for inspection.

<pre><code>dictvar = countWords(listoftweets)
</code></pre>

creating a definition will allow you to place this inside of its own file that you can always import use in the future.

I have a list of tweets that is grouped into chunks of tweets within the list like so: 

<pre><code>[[tweet1, tweet2, tweet3],[tweet4,tweet5,tweet6],[tweet7, tweet8, tweet9]]
</code></pre>

I want to count the number of occurences of each word within each subgroup. To do this, I need to split each tweet into individual words. I want to use something similar to str.split(' '), but I receive an error: 

<pre><code>AttributeError: 'list' object has no attribute 'split' 
</code></pre>

Is there a way to split each tweet into its individual words? The result should looks something like: 

<pre><code>[['word1', 'word2', 'word3', 'word2', 'word2'],['word1', 'word1', 'word3', 'word4', 'word5'],['word1', 'word3', 'word3', 'word5', 'word6']]
</code></pre>

Manipulating strings in python list

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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我有一个tweet列表，它被分组为列表中的tweet块，如下所示：[[tweet1, tweet2, tweet3],[tweet4,tweet5,tweet6],[tweet7, tweet8, tweet9]]我想统计每个单词在每个子组中出现的次数。要做到这一点，我需要将每个tweet分成单独的单词。我想使用类似于...

问在python列表中操作字符串
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