blocks|key|4193215|text|只需使用一个范围：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4193216|for+i+in+range(0,+size,+n):
++++print+(i)|code-block|syntax|javascript|4193217|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 use a range:

<pre><code>for i in range(0, size, n):
 print (i)
</code></pre>

blocks|key|4193267|text|我不确定哪种值是n，但通常有这样的方式：(对我来说，n是一个列表)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4193268|for+index,+item+in+enumerate(n):
+++if+index+%25+nth+==+0:+#+If+the+output+is+not+like+you+want,+try+doing+(index+%2B+1),+or+enumerate(n,+start=1).
+++++++print(item)|code-block|syntax|javascript|4193269|另一种方式可能是：|4193270|for+index+in+range(0,+len(n),+nth):+#+Only+work+with+sequences
+++print(n[index])+#+If+the+output+is+not+like+you+want,+try+doing+n[index+%2B+1]|4193271|或者：|4193272|for+item+in+n[::nth]:+#+Low+perfomance+and+hight+memory+consumption+warning!!+Only+work+with+sequences
++++print(item)|4193273|甚至你也可以把第一个和最后一个结合起来：|4193274|for+i,+item+in+list(enumerate(n))[::nth]:+#+Huge+low+perfomance+warning!!!
++++print(item)|4193275|但我不确定这是否有优势。|4193276|此外，如果您愿意创建一个函数，则可以执行类似于enumerate函数的操作：|4193277|def+myEnumerate(sequence,+start=0,+jump=1):
++++n+=+start
++++j+=+start+//+Or+j+=+0,+that+is+your+decision.
++++for+elem+in+sequence:
++++++++if+j+%25+jump+==+0:
++++++++++++yield+n,+elem
++++++++++++n+%2B=+1
++++++++j+%2B=+1

for+index,+item+in+myEnumerate(n,+jump=1):
++++print(item)|4193278|就我个人而言，我不会做最后一个。我不知道为什么，但这是一种感觉。|4193279|性能测试|4193280|n+=+'a+b+c+d+e+f+g+h+i+j+k+l+m+n+ñ+o+p+q+r+s+t+u+v+w+x+y+z+1+2+3+4+5+6+7+8+9+!+"+·+$+%25+&+/+(+)+=+?+¿+Ç+ç+}+{+[+]+;+:+_+¨+%5E+*+`+%2B+´+-+.+,+º+ª+\+/'.split("+")
nth+=+3++++
def+a():
++++for+i,+item+in+enumerate(n):
+++++++if+i+%25+nth+==+0:
+++++++++++item+++++++
def+b():
++++for+item+in+range(0,+len(n),+nth):
+++++++n[item]+++++++++++
def+c():
++++for+item+in+n[::nth]:
++++++++item++++
def+d():
++++for+i,+item+in+list(enumerate(n))[::nth]:
+++++++if+i+%25+nth+==+0:
+++++++++++item++++
def+enumerates(sequence,+start=0,+jump=1):
++++n+=+start
++++j+=+start
++++for+elem+in+sequence:
++++++++if+j+%25+jump+==+0:
++++++++++++yield+n,+elem
++++++++++++n+%2B=+1
++++++++j+%2B=+1++++++++++++
def+e():
++++for+i,+item+in+enumerates(n,+jump=+nth):
++++++++item++++
if+__name__+==+'__main__':
++++import+timeit
++++print(timeit.timeit("a()",+setup="from+__main__+import+a"))+#+10.556324407152305
++++print(timeit.timeit("b()",+setup="from+__main__+import+b"))+#+2.7166204783010137
++++print(timeit.timeit("c()",+setup="from+__main__+import+c"))+#+1.0285353306076601
++++print(timeit.timeit("d()",+setup="from+__main__+import+d"))+#+8.283859051918608
++++print(timeit.timeit("e()",+setup="from+__main__+import+e"))+#+14.91601851631981|4193281|但如果你真的想获得更高的性能，你应该阅读@Martijn+Pieters+answer。|4193282|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/50786774/7655838^0|8|1|Q|1|0|0|0|0|0|0|0|0|0|N|9|0|0|0|0|0|K|N|0|0^^$0|@$1|2|3|4|5|6|7|1I|8|@$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1N|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1O|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1P|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1Q|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1R|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1S|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|1T|8|@]|D|@]|E|$I|J]]|$1|W|3|X|5|6|7|1U|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|6|7|1V|8|@$9|1W|A|1X|B|C]]|D|@]|E|$]]|$1|10|3|11|5|H|7|1Y|8|@]|D|@]|E|$I|J]]|$1|12|3|13|5|6|7|1Z|8|@]|D|@]|E|$]]|$1|14|3|15|5|6|7|20|8|@]|D|@]|E|$]]|$1|16|3|17|5|H|7|21|8|@]|D|@]|E|$I|J]]|$1|18|3|19|5|6|7|22|8|@]|D|@$9|23|A|24|1|25]]|E|$]]|$1|1A|3|-4|5|6|7|26|8|@]|D|@]|E|$]]]|1B|$1C|$5|1D|1E|1F|E|$1G|1H]]]]

I am not sure wich kind of value is <code>n</code> but usually there are this ways: (for me, <code>n</code> is a list)

<pre><code>for index, item in enumerate(n):
 if index % nth == 0: # If the output is not like you want, try doing (index + 1), or enumerate(n, start=1).
 print(item)
</code></pre>

Other way could be:

<pre><code>for index in range(0, len(n), nth): # Only work with sequences
 print(n[index]) # If the output is not like you want, try doing n[index + 1]
</code></pre>

Or:

<pre><code>for item in n[::nth]: # Low perfomance and hight memory consumption warning!! Only work with sequences
 print(item)
</code></pre>

Even you can combine the first one with the last one:

<pre><code>for i, item in list(enumerate(n))[::nth]: # Huge low perfomance warning!!!
 print(item)
</code></pre>

But I'm not sure if that has an advantage...

Also, if you are willing to make a function, you could do something similar to the <code>enumerate</code> function:

<pre><code>def myEnumerate(sequence, start=0, jump=1):
 n = start
 j = start // Or j = 0, that is your decision.
 for elem in sequence:
 if j % jump == 0:
 yield n, elem
 n += 1
 j += 1

for index, item in myEnumerate(n, jump=1):
 print(item)
</code></pre>

Personally, I wouldn't do this last one. I'm not sure why but it's a feeling.

<h3>Perfomance test</h3>

<pre><code>n = 'a b c d e f g h i j k l m n ñ o p q r s t u v w x y z 1 2 3 4 5 6 7 8 9 ! " · $ % &amp; / ( ) = ? ¿ Ç ç } { [ ] ; : _ ¨ ^ * ` + ´ - . , º ª \ /'.split(" ")
nth = 3 
def a():
 for i, item in enumerate(n):
 if i % nth == 0:
 item 
def b():
 for item in range(0, len(n), nth):
 n[item] 
def c():
 for item in n[::nth]:
 item 
def d():
 for i, item in list(enumerate(n))[::nth]:
 if i % nth == 0:
 item 
def enumerates(sequence, start=0, jump=1):
 n = start
 j = start
 for elem in sequence:
 if j % jump == 0:
 yield n, elem
 n += 1
 j += 1 
def e():
 for i, item in enumerates(n, jump= nth):
 item 
if __name__ == '__main__':
 import timeit
 print(timeit.timeit("a()", setup="from __main__ import a")) # 10.556324407152305
 print(timeit.timeit("b()", setup="from __main__ import b")) # 2.7166204783010137
 print(timeit.timeit("c()", setup="from __main__ import c")) # 1.0285353306076601
 print(timeit.timeit("d()", setup="from __main__ import d")) # 8.283859051918608
 print(timeit.timeit("e()", setup="from __main__ import e")) # 14.91601851631981
</code></pre>

But if you are really looking for perfomance you should read <a href="https://stackoverflow.com/a/50786774/7655838">@Martijn Pieters answer</a>.

How can I create a for loop with a counter? I have a list, and I want to read an element after each <code>n</code> elements. I'd initially done this

<pre><code>for i in enumerate(n):
 print(i)
</code></pre>

But as expected it prints every element instead of every nth element, what would be the Python way of solving this?

for loop in python with counter?

如何创建一个带有计数器的for循环？我有一个列表，我想在每个n元素之后读取一个元素。我最初是这么做的for i in enumerate(n):    print(i)但是正如预期的那样，它打印每个元素，而不是每第n个元素，那么Python解决这个问题的方法是什么？

问使用计数器的python中的for循环？
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