blocks|key|1900230|text|使用zip|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1900231|[m%2Bstr(n)+for+m,n+in+zip(b,a)]|code-block|syntax|javascript|1900232|输出|1900233|['asp10',+'asp11',+'asp15',+'asp16',+'asp210',+'asp211']|1900234|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|$]]

Using zip

<pre><code>[m+str(n) for m,n in zip(b,a)]
</code></pre>

output

<pre><code>['asp10', 'asp11', 'asp15', 'asp16', 'asp210', 'asp211']
</code></pre>

blocks|key|1686862|text|其他解决方案(更喜欢printf格式样式而不是.format()用法)，它也更小：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1686863|>>>+["%25s%2502d"+%25+t+for+t+in+zip(b,+a)]
['asp100',+'asp101',+'asp105',+'asp106',+'asp210',+'asp211']|code-block|syntax|javascript|1686864|entityMap^0|N|9|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|$]]

Other solution (preferring printf formating style over <code>.format()</code> usage), it's also smaller:

<pre><code>&gt;&gt;&gt; ["%s%02d" % t for t in zip(b, a)]
['asp100', 'asp101', 'asp105', 'asp106', 'asp210', 'asp211']
</code></pre>

blocks|key|1900090|text|比用map和zip做得更好：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1900091|map(lambda+(x,y):+x%2By,+zip(list1,+list2))|code-block|syntax|javascript|1900092|示例：|1900093|In+[1]:+map(lambda+(x,y):+x%2By,+zip([1,2,3,4],[4,5,6,7]))
Out[1]:+[5,+7,+9,+11]|1900094|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|$]]

Than can be done elegantly with map and zip:

<pre><code>map(lambda (x,y): x+y, zip(list1, list2))
</code></pre>

Example: 

<pre><code>In [1]: map(lambda (x,y): x+y, zip([1,2,3,4],[4,5,6,7]))
Out[1]: [5, 7, 9, 11]
</code></pre>

blocks|key|1900358|text|输入：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1900359|a+=+[0,+1,+5,+6,+10,+11]+
b+=+['asp1',+'asp1',+'asp1',+'asp1',+'asp2',+'asp2']

concat_func+=+lambda+x,y:+x+%2B+""+%2B+str(y)

list(map(concat_func,b,a))+#+list+the+map+function|code-block|syntax|javascript|1900360|输出：|1900361|['asp10',+'asp11',+'asp15',+'asp16',+'asp210',+'asp211']|1900362|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|$]]

inputs:

<pre><code>a = [0, 1, 5, 6, 10, 11] 
b = ['asp1', 'asp1', 'asp1', 'asp1', 'asp2', 'asp2']

concat_func = lambda x,y: x + "" + str(y)

list(map(concat_func,b,a)) # list the map function
</code></pre>

output: 

<pre><code>['asp10', 'asp11', 'asp15', 'asp16', 'asp210', 'asp211']
</code></pre>

blocks|key|1900364|text|如果你想连接任意数量的列表，你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1900365|In+[1]:+lists+=+[["a",+"b",+"c"],+["m",+"n",+"o"],+["p",+"q",+"r"]]+#+Or+more

In+[2]:+lists
Out[2]:+[['a',+'b',+'c'],+['m',+'n',+'o'],+['p',+'q',+'r']]++++

In+[4]:+list(map("".join,+zip(*lists)))
Out[4]:+['amp',+'bnq',+'cor']|code-block|syntax|javascript|1900366|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|$]]

If you wanted to concatenate arbitrary number of lists, you could do this:

<pre><code>In [1]: lists = [["a", "b", "c"], ["m", "n", "o"], ["p", "q", "r"]] # Or more

In [2]: lists
Out[2]: [['a', 'b', 'c'], ['m', 'n', 'o'], ['p', 'q', 'r']] 

In [4]: list(map("".join, zip(*lists)))
Out[4]: ['amp', 'bnq', 'cor']
</code></pre>

blocks|key|1686691|text|不是使用zip。我不知道，我想这是最明显的方法。也许我只是先学习了C语言:)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1686692|c=[]
for+i+in+xrange(len(a)):
++++c.append("%25s%2502d"+%25+(b[i],a[i]))|code-block|syntax|javascript|1686693|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|$]]

not using zip. I dunno, I think this is the obvious way to do it. Maybe I just learnt C first :)

<pre><code>c=[]
for i in xrange(len(a)):
 c.append("%s%02d" % (b[i],a[i]))
</code></pre>

blocks|key|1686777|text|b+=+['asp1',+'asp1',+'asp1',+'asp1',+'asp2',+'asp2']
aa+=+[0,+1,+5,+6,+10,+11]
new_list+=[]
if+len(aa)+!=+len(b):
+++++print+'list+length+mismatch'
else:
++++for+each+in+range(0,len(aa)):
++++++++new_list.append(b[each]+%2B+str(aa[each]))
print+new_list|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1686778|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>b = ['asp1', 'asp1', 'asp1', 'asp1', 'asp2', 'asp2']
aa = [0, 1, 5, 6, 10, 11]
new_list =[]
if len(aa) != len(b):
 print 'list length mismatch'
else:
 for each in range(0,len(aa)):
 new_list.append(b[each] + str(aa[each]))
print new_list
</code></pre>

blocks|key|1686895|text|我最终使用了一个临时的DataFrame，它可读性好，速度快：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1686896|a+=+["a",+"b",+"c"]
b+=+["1",+"2",+"3"]

df+=+pd.DataFrame({"a":+a,+"b":+b})
df["c"]+=+df.a+%2B+df.b
result+=+df.c.values|code-block|syntax|javascript|1686897|输出：|1686898|$+result+
["a1",+"b2",+"c3"]|1686899|在幕后，DataFrames使用numpy，因此结果是高效的。|1686900|1686901|和函数是一样的：|1686902|import+pandas+as+pd
from+typing+import+List
def+_elementwise_concat(self,+a:+List[str],+b:+List[str])+->+List[str]:
++++"""
++++Elementwise+concatenate.
++++:param+a:+List+of+strings.
++++:param+b:+List+of+strings.
++++:return:+List,+same+length,+strings+concatenated.
++++"""
++++df+=+pd.DataFrame({"a":+a,+"b":+b})
++++df["c"]+=+df.a+%2B+df.b
++++return+df.c.values|1686903|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|U|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|V|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|W|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|X|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|Z|8|@]|9|@]|A|$]]|$1|P|3|Q|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|R|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|S|$]]

I ended up using a temporary DataFrame, it's readable and fast:
<pre><code>a = [&quot;a&quot;, &quot;b&quot;, &quot;c&quot;]
b = [&quot;1&quot;, &quot;2&quot;, &quot;3&quot;]

df = pd.DataFrame({&quot;a&quot;: a, &quot;b&quot;: b})
df[&quot;c&quot;] = df.a + df.b
result = df.c.values
</code></pre>
Output:
<pre><code>$ result 
[&quot;a1&quot;, &quot;b2&quot;, &quot;c3&quot;]
</code></pre>
Behind the scenes, DataFrames use numpy, so the result is efficient.
<hr />
And the same thing as a function:
<pre><code>import pandas as pd
from typing import List
def _elementwise_concat(self, a: List[str], b: List[str]) -&gt; List[str]:
 &quot;&quot;&quot;
 Elementwise concatenate.
 :param a: List of strings.
 :param b: List of strings.
 :return: List, same length, strings concatenated.
 &quot;&quot;&quot;
 df = pd.DataFrame({&quot;a&quot;: a, &quot;b&quot;: b})
 df[&quot;c&quot;] = df.a + df.b
 return df.c.values
</code></pre>

blocks|key|1900404|text|列出理解/+zip()+/使用zfill()进行格式化。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1900405|print+([y%2Bstr(x).zfill(2)+for+x,y+in+zip(a,b)])|code-block|syntax|javascript|1900406|输出：|1900407|['asp100',+'asp101',+'asp105',+'asp106',+'asp210',+'asp211']|1900408|entityMap^0|6|5|F|7|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

List comprehension / <code>zip()</code> / use of <code>zfill()</code> to format.
<pre><code>print ([y+str(x).zfill(2) for x,y in zip(a,b)])
</code></pre>
Output:
<pre><code>['asp100', 'asp101', 'asp105', 'asp106', 'asp210', 'asp211']
</code></pre>

blocks|key|1900410|text|使用lambda和format|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1900411|b+=+['asp1',+'asp1',+'asp1',+'asp1',+'asp2',+'asp2']
a+=+[0,+1,+5,+6,+10,+11]

list(map(lambda+x:+x[0]+%2B+"{0:0=2d}".format(x[1]),+zip(b,+a)))|code-block|syntax|javascript|1900412|输出：|1900413|['asp100',+'asp101',+'asp105',+'asp106',+'asp210',+'asp211']|1900414|entityMap^0|2|6|9|6|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

Using <code>lambda</code> and <code>format</code>:
<pre><code>b = ['asp1', 'asp1', 'asp1', 'asp1', 'asp2', 'asp2']
a = [0, 1, 5, 6, 10, 11]

list(map(lambda x: x[0] + &quot;{0:0=2d}&quot;.format(x[1]), zip(b, a)))
</code></pre>
Out:
<pre><code>['asp100', 'asp101', 'asp105', 'asp106', 'asp210', 'asp211']
</code></pre>

I have two lists and I want to concatenate them element-wise. One of the list is subjected to string-formatting before concatenation. 

For example :

<pre><code>a = [0, 1, 5, 6, 10, 11] 
b = ['asp1', 'asp1', 'asp1', 'asp1', 'asp2', 'asp2']
</code></pre>

In this case, <code>a</code> is subjected to string-formatting. That is, new <code>a</code> or <code>aa</code> should be :

<pre><code>aa = [00, 01, 05, 06, 10, 11]
</code></pre>

Final output should be :

<pre><code>c = ['asp100', 'asp101', 'asp105', 'asp106', 'asp210', 'asp211']
</code></pre>

Can somebody please tell me how to do that?

How to concatenate element-wise two lists in Python

我有两个列表，我想按元素将它们连接起来。其中一个列表在连接之前要进行字符串格式设置。例如：a = [0, 1, 5, 6, 10, 11] b = ['asp1', 'asp1', 'asp1', 'asp1', 'asp2', 'asp2']在这种情况下，a需要进行字符串格式化。也就是说，新的a或aa应该是：aa = [00, 01, 05, 06, 10, 11]最终输出应为：c = ['as

问如何在Python中以元素方式连接两个列表
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