blocks|key|1145879|text|要做到这一点，最简单的方法是列表理解：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1145880|[s+%2B+mystring+for+s+in+mylist]|code-block|syntax|javascript|1145881|请注意，我避免使用像list这样的内置名称，因为那样会隐藏或隐藏内置名称，这非常不好。|offset|length|style|CODE|1145882|此外，如果您实际上不需要列表，但只需要一个迭代器，那么生成器表达式可能会更有效(尽管这在短列表中可能并不重要)：|1145883|(s+%2B+mystring+for+s+in+mylist)|1145884|这些都是非常强大、灵活和简洁的。每一个优秀的python程序员都应该学会使用它们。|1145885|entityMap^0|0|0|A|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

The simplest way to do this is with a list comprehension:

<pre><code>[s + mystring for s in mylist]
</code></pre>

Notice that I avoided using builtin names like <code>list</code> because that shadows or hides the builtin names, which is very much not good.

Also, if you do not actually need a list, but just need an iterator, a generator expression can be more efficient (although it does not likely matter on short lists):

<pre><code>(s + mystring for s in mylist)
</code></pre>

These are very powerful, flexible, and concise. Every good python programmer should learn to wield them.

blocks|key|1145943|text|list2+=+['%25sbar'+%25+(x,)+for+x+in+list]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1145944|并且不要使用list作为名称；它隐藏了内置类型。|unstyled|offset|length|style|CODE|1145945|entityMap^0|0|6|4|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>list2 = ['%sbar' % (x,) for x in list]
</code></pre>

And don't use <code>list</code> as a name; it shadows the built-in type.

blocks|key|1145998|text|new_list+=+[word_in_list+%2B+end_string+for+word_in_list+in+old_list]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1145999|使用像"list“这样的名称作为你的变量名是不好的，因为它会覆盖/覆盖内置的东西。|unstyled|1146000|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>new_list = [word_in_list + end_string for word_in_list in old_list]
</code></pre>

Using names such as "list" for your variable names is bad since it will overwrite/override the builtins.

blocks|key|1146020|text|my_list+=+['foo',+'fob',+'faz',+'funk']
string+=+'bar'
my_new_list+=+[x+%2B+string+for+x+in+my_list]
print+my_new_list|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1146021|这将打印以下内容：|unstyled|1146022|['foobar',+'fobbar',+'fazbar',+'funkbar']|1146023|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>my_list = ['foo', 'fob', 'faz', 'funk']
string = 'bar'
my_new_list = [x + string for x in my_list]
print my_new_list
</code></pre>

This will print:

<pre><code>['foobar', 'fobbar', 'fazbar', 'funkbar']
</code></pre>

blocks|key|1361089|text|在我看来，map似乎是适合这项工作的工具。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1361090|my_list+=+['foo',+'fob',+'faz',+'funk']
string+=+'bar'
list2+=+list(map(lambda+orig_string:+orig_string+%2B+string,+my_list))|code-block|syntax|javascript|1361091|有关map的更多示例，请参阅函数式编程工具上的this+section。|1361092|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/2/tutorial/datastructures.html#functional-programming-tools^0|5|3|0|0|2|3|N|C|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|M|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

<code>map</code> seems like the right tool for the job to me.

<pre><code>my_list = ['foo', 'fob', 'faz', 'funk']
string = 'bar'
list2 = list(map(lambda orig_string: orig_string + string, my_list))
</code></pre>

See <a href="http://docs.python.org/2/tutorial/datastructures.html#functional-programming-tools" rel="nofollow noreferrer">this section</a> on functional programming tools for more examples of <code>map</code>.

blocks|key|1361145|text|你可以在python的map中使用lambda。写了一个格雷码生成器。https://github.com/rdm750/rdm750.github.io/blob/master/python/gray_code_generator.py+#您的代码放在这里“”n-1位代码，每个字前面加上0，然后是反顺序的n-1位代码，每个字前面加1。“”“|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1361146|++++def+graycode(n):
++++++++if+n==1:
++++++++++++return+['0','1']
++++++++else:
++++++++++++nbit=map(lambda+x:'0'%2Bx,graycode(n-1))%2Bmap(lambda+x:'1'%2Bx,graycode(n-1)[::-1])
++++++++++++return+nbit

++++for+i+in+xrange(1,7):
++++++++print+map(int,graycode(i))|code-block|syntax|javascript|1361147|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/rdm750/rdm750.github.io/blob/master/python/gray_code_generator.py^0|Z|2C|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

you can use lambda inside map in python. wrote a gray codes generator.
<a href="https://github.com/rdm750/rdm750.github.io/blob/master/python/gray_code_generator.py" rel="nofollow">https://github.com/rdm750/rdm750.github.io/blob/master/python/gray_code_generator.py</a>
 # your code goes here
 '''
 the n-1 bit code, with 0 prepended to each word, followed by
 the n-1 bit code in reverse order, with 1 prepended to each word.
 '''

<pre><code> def graycode(n):
 if n==1:
 return ['0','1']
 else:
 nbit=map(lambda x:'0'+x,graycode(n-1))+map(lambda x:'1'+x,graycode(n-1)[::-1])
 return nbit

 for i in xrange(1,7):
 print map(int,graycode(i))
</code></pre>

blocks|key|1361164|text|这是一个使用pandas的简单答案。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1361165|import+pandas+as+pd
list1+=+['foo',+'fob',+'faz',+'funk']
string+=+'bar'

list2+=+(pd.Series(list1)+%2B+string).tolist()
list2
#+['foobar',+'fobbar',+'fazbar',+'funkbar']|code-block|syntax|javascript|1361166|entityMap^0|6|6|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|$]]

Here is a simple answer using <code>pandas</code>.

<pre class="lang-python prettyprint-override"><code>import pandas as pd
list1 = ['foo', 'fob', 'faz', 'funk']
string = 'bar'

list2 = (pd.Series(list1) + string).tolist()
list2
# ['foobar', 'fobbar', 'fazbar', 'funkbar']
</code></pre>

blocks|key|1146184|text|以防万一|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1146185|list+=+['foo',+'fob',+'faz',+'funk']
string+=+'bar'
for+i+in+range(len(list)):
++++list[i]+%2B=+string
print(list)|code-block|syntax|javascript|1146186|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Just in case

<pre><code>list = ['foo', 'fob', 'faz', 'funk']
string = 'bar'
for i in range(len(list)):
 list[i] += string
print(list)
</code></pre>

blocks|key|1361207|text|结合使用map和format|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1361208|>>>+list(map('{0}bar'.format,++['foo',+'fob',+'faz',+'funk']))
['foobar',+'fobbar',+'fazbar',+'funkbar']|code-block|syntax|javascript|1361209|entityMap^0|4|3|8|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Combining <code>map</code> and <code>format</code>:
<pre><code>&gt;&gt;&gt; list(map('{0}bar'.format, ['foo', 'fob', 'faz', 'funk']))
['foobar', 'fobbar', 'fazbar', 'funkbar']
</code></pre>

I am trying to take one string, and append it to every string contained in a list, and then have a new list with the completed strings. Example:
<pre><code>list1 = ['foo', 'fob', 'faz', 'funk']
string = 'bar'

*magic*

list2 = ['foobar', 'fobbar', 'fazbar', 'funkbar']
</code></pre>
I tried for loops, and an attempt at list comprehension, but it was garbage. As always, any help, much appreciated.

Appending the same string to a list of strings in Python

我尝试获取一个字符串，并将其附加到列表中包含的每个字符串，然后创建一个包含完整字符串的新列表。示例：list1 = ['foo', 'fob', 'faz', 'funk']string = 'bar'*magic*list2 = ['foobar', 'fobbar', 'fazbar', 'funkbar']我尝试...

问将相同的字符串追加到Python中的字符串列表中
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