blocks|key|5916357|text|mystring+=+"You+have+15+new+messages+and+the+size+is+32000"
parts+=+mystring.split('+')
message_count+=+int(parts[2])
message_size+=+int(parts[9])|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|5916358|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>mystring = "You have 15 new messages and the size is 32000"
parts = mystring.split(' ')
message_count = int(parts[2])
message_size = int(parts[9])
</code></pre>

blocks|key|1815288|text|看起来您正在匹配程序输出或日志文件中的内容。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1815289|在这种情况下，您希望进行足够的匹配，以便有信心匹配正确的内容，但不要太多，以至于如果输出稍有变化，程序就会出错。|1815290|正则表达式在这种情况下工作得很好，例如|1815291|>>>+import+re
>>>+mystring+=+"You+have+15+new+messages+and+the+size+is+32000"
>>>+match+=+re.search(r"(\d%2B).*?messages.*?size.*?(\d%2B)",+mystring)
>>>+if+not+match:+print+"log+line+didn't+match"
...+
>>>+messages,+size+=+map(int,+match.groups())
>>>+messages
15
>>>+size
32000|code-block|syntax|javascript|1815292|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|P|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

It looks like you are matching something from program output or a log file.

In this case you want to match enough so you have confidence you are matching the right thing, but not so much that if the output changes a little bit your program goes wrong.

Regular expressions work well in this case, eg

<pre><code>&gt;&gt;&gt; import re
&gt;&gt;&gt; mystring = "You have 15 new messages and the size is 32000"
&gt;&gt;&gt; match = re.search(r"(\d+).*?messages.*?size.*?(\d+)", mystring)
&gt;&gt;&gt; if not match: print "log line didn't match"
... 
&gt;&gt;&gt; messages, size = map(int, match.groups())
&gt;&gt;&gt; messages
15
&gt;&gt;&gt; size
32000
</code></pre>

blocks|key|2388421|text|mystring+=+"You+have+15+new+messages+and+the+size+is+32000"
+
print+mystring.split("+")[2]++#Prints+the+3rd+word

print+mystring.split("+")[9]+#Prints+the+10th+word|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2388422|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>mystring = &quot;You have 15 new messages and the size is 32000&quot;
 
print mystring.split(&quot; &quot;)[2] #Prints the 3rd word

print mystring.split(&quot; &quot;)[9] #Prints the 10th word
</code></pre>

blocks|key|4500915|text|下面的函数实现了这个功能：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4500916|def+giveme(s,+words=()):
++++lista+=+s.split()++++
++++return+[lista[item-1]+for+item+in+words]+++

mystring+=+"You+have+15+new+messages+and+the+size+is+32000"
position+=+(3,+10)
print+giveme(mystring,+position)

it+prints+->+['15',+'32000']|code-block|syntax|javascript|4500917|Ignacio指出的替代方法非常简单：|4500918|import+operator

mystring+=+"You+have+15+new+messages+and+the+size+is+32000"
position+=+(2,+9)

lista+=+mystring.split()
f+=+operator.itemgetter(*position)
print+f(lista)

it+prints+->+['15',+'32000']|4500919|4500920|operator.itemgetter()+...|offset|length|style|CODE|4500921|4500922|返回一个可调用的对象，该对象从其操作数中获取给定项。|blockquote|4500923|4500924|在f+=+itemgetter(2)之后，调用f(r)返回r2。|4500925|在g+=+itemgetter(2,5,3)之后，调用g(r)返回(r2，r5，r3)|4500926|4500927|请注意，现在position中的位置应从0开始计数，以允许直接使用*position参数|BOLD|4500928|entityMap^0|0|0|0|0|0|0|L|0|0|0|0|1|H|N|4|0|1|L|R|4|0|0|6|8|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|18|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|1B|8|@$N|1C|O|1D|P|Q]]|9|@]|A|$]]|$1|R|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]|$1|S|3|T|5|U|7|1F|8|@]|9|@]|A|$]]|$1|V|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|1H|8|@$N|1I|O|1J|P|Q]|$N|1K|O|1L|P|Q]]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|1M|8|@$N|1N|O|1O|P|Q]|$N|1P|O|1Q|P|Q]]|9|@]|A|$]]|$1|10|3|-4|5|6|7|1R|8|@]|9|@]|A|$]]|$1|11|3|12|5|6|7|1S|8|@$N|1T|O|1U|P|13]]|9|@]|A|$]]|$1|14|3|-4|5|6|7|1V|8|@]|9|@]|A|$]]]|15|$]]

This function does the trick:

<pre><code>def giveme(s, words=()):
 lista = s.split() 
 return [lista[item-1] for item in words] 

mystring = "You have 15 new messages and the size is 32000"
position = (3, 10)
print giveme(mystring, position)

it prints -&gt; ['15', '32000']
</code></pre>

The alternative indicated by Ignacio is very clean:

<pre><code>import operator

mystring = "You have 15 new messages and the size is 32000"
position = (2, 9)

lista = mystring.split()
f = operator.itemgetter(*position)
print f(lista)

it prints -&gt; ['15', '32000']
</code></pre>

<hr>

<code>operator.itemgetter()</code> ... 

<blockquote>
 Return a callable object that fetches
 the given item(s) from its operand.
 
 After, <code>f = itemgetter(2)</code>, the call <code>f(r)</code>
 returns r[2].
 
 After, <code>g = itemgetter(2,5,3)</code>, the call <code>g(r)</code>
 returns (r[2], r[5], r[3])
</blockquote>

Note that now positions in position should be counted from 0 to allow using directly the *position argument

blocks|key|1811549|text|看看str.split()吧。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1811550|或者，如果您正在寻找某些东西，您可以尝试使用正则表达式；这甚至可以应对填充词的变化。但是，如果您只关心字符串中的单词位置，那么拆分并打印出结果列表中的某些元素将是最直接的。|1811551|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/stdtypes.html#str.split^0|2|B|2|B|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@$9|S|A|T|1|U]]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

Take a look at <a href="http://docs.python.org/library/stdtypes.html#str.split" rel="nofollow noreferrer"><code>str.split()</code></a>.

Alternatively, if you're looking for certain things you might try regular expressions instead; that could cope with the filler words changing, even. But if all you care about is word position within the string, then splitting and printing out certain elements of the resulting list would be the most straightforward.

blocks|key|1811594|text|这个怎么样：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1811595|import+re

tst_str+=+"You+have+15+new+messages+and+the+size+is+32000"
items+=+re.findall(r"+*\d%2B+*",tst_str)
for+item+in+items:
++++print(item)|code-block|syntax|javascript|1811596|结果：|1811597|+15+
+32000|1811598|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

How about this one:

<pre><code>import re

tst_str = "You have 15 new messages and the size is 32000"
items = re.findall(r" *\d+ *",tst_str)
for item in items:
 print(item)
</code></pre>

Result:

<pre><code> 15 
 32000
</code></pre>

How can i print only certain words from a string in python ?
lets say i want to print only the 3rd word (which is a number) and the 10th one

while the text length may be different each time 

<code>mystring = "You have 15 new messages and the size is 32000"</code>

thanks.

Print one word from a string 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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如何在python中仅打印字符串中的特定单词？假设我只想打印第三个单词(这是一个数字)和第十个单词而文本长度每次都可能不同mystring = "You have 15 new messages and the size is 32000"谢谢。

问在python中打印字符串中的一个单词
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