blocks|key|2983528|text|Python切片语法支持第三个可选参数，即步骤参数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2983529|def+oddIndexes(string):
++++return+string[1::2]

>>>+oddIndexes('102030')
'000'|code-block|syntax|javascript|2983530|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|$]]

Python slicing syntax supports an optional third argument, a step argument:

<pre><code>def oddIndexes(string):
 return string[1::2]

&gt;&gt;&gt; oddIndexes('102030')
'000'
</code></pre>

blocks|key|1254541|text|编辑：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|1254542|否则，如果您从给定的字符串开始，那么只需将其切片如下：|1254543|a+=+a[1::2]|code-block|syntax|javascript|1254544|第一个元素1意味着要从其第二个元素开始对元素进行切片。|ordered-list-item|CODE|1254545|最后一个元素2是间隔(每两个元素)|1254546|空中间元素实际上是索引上限(排它)。如果你不放任何东西就意味着你想要所有的东西。|1254547|原版：|1254548|参考您先前的问题，在此：|1254549|如何将列表元素变成字符串？|1254550|从拥有a+=+[10,+20,+30]开始，|1254551|为了从每个列表项的第二个元素中获取给定整数列表a的联接字符串，可以使用[1::]+(索引1表示从第二个元素开始)将列表的每个项分割到第二个元素开始，如下所示：|1254552|a+=+[10,+20,+30]
a+=+[''.join(str(x)[1::]+for+x+in+a)]

print(a)|1254553|结果：|1254554|['000']|1254555|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/43904854/how-to-make-list-elements-into-string^0|0|3|0|0|0|5|1|0|6|1|0|0|0|3|0|0|0|D|0|0|3|G|0|N|1|Z|5|18|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1I|8|@$9|1J|A|1K|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1L|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|1M|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|O|7|1N|8|@$9|1O|A|1P|B|P]]|D|@]|E|$]]|$1|Q|3|R|5|O|7|1Q|8|@$9|1R|A|1S|B|P]]|D|@]|E|$]]|$1|S|3|T|5|O|7|1T|8|@]|D|@]|E|$]]|$1|U|3|V|5|6|7|1U|8|@$9|1V|A|1W|B|C]]|D|@]|E|$]]|$1|W|3|X|5|6|7|1X|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|6|7|1Y|8|@]|D|@$9|1Z|A|20|1|21]]|E|$]]|$1|10|3|11|5|6|7|22|8|@$9|23|A|24|B|P]]|D|@]|E|$]]|$1|12|3|13|5|6|7|25|8|@$9|26|A|27|B|P]|$9|28|A|29|B|P]|$9|2A|A|2B|B|P]]|D|@]|E|$]]|$1|14|3|15|5|J|7|2C|8|@]|D|@]|E|$K|L]]|$1|16|3|17|5|6|7|2D|8|@]|D|@]|E|$]]|$1|18|3|19|5|J|7|2E|8|@]|D|@]|E|$K|L]]|$1|1A|3|-4|5|6|7|2F|8|@]|D|@]|E|$]]]|1B|$1C|$5|1D|1E|1F|E|$1G|1H]]]]

Edit:

Else, if you start from the given string, then you should simply slice it like this:

<pre><code>a = a[1::2]
</code></pre>

<ol>
<li>First element <code>1</code> means you want to slice the element starting from its second element.</li>
<li>The last element <code>2</code> is the interval (every two element)</li>
<li>The empty middle element is actually the upper index limit (exclusive). If you don't put anything means you want all items.</li>
</ol>

Original:

Refering from your previous question here:

<a href="https://stackoverflow.com/questions/43904854/how-to-make-list-elements-into-string">How to make list elements into string?</a>

Started from having <code>a = [10, 20, 30]</code>,

In order to get the joined string for the given integer list <code>a</code> from the second element per list item onwards, you could slice each item of the list to from the second element onwards by using <code>[1::]</code> (index <code>1</code> means starting from the second element) like this:

<pre><code>a = [10, 20, 30]
a = [''.join(str(x)[1::] for x in a)]

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

Result:

<pre><code>['000']
</code></pre>

blocks|key|1271165|text|只需做以下几点：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1271166|l[1:][::2]|code-block|syntax|javascript|1271167|它所做的事情基本上与：|1271168|l[1::2]|1271169|>>>+a=+'102030'
>>>+a[1:][::2]
'000'
>>>+a[1::2]
'000'
>>>|1271170|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]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Just do the following:

<pre><code>l[1:][::2]
</code></pre>

Which does essentially the same thing as:

<pre><code>l[1::2]
</code></pre>

<hr>

<pre><code>&gt;&gt;&gt; a= '102030'
&gt;&gt;&gt; a[1:][::2]
'000'
&gt;&gt;&gt; a[1::2]
'000'
&gt;&gt;&gt;
</code></pre>

blocks|key|2983626|text|首先，您只能对某些序列对象进行切片，如列表和字符串。你有整数。不过，你仍然可以得到你想要的结果。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2983627|首先，将您的数字转换为字符串：|2983628|a+=+102030
str_a+str(a)|code-block|syntax|javascript|2983629|然后，您可以使用普通的切片语法，所以您可以使用开始、结束、步骤大小来完成这个任务。|offset|length|2983630|a+=+int(str_a[1::2])
#+a+==+0
#+but+str_a[1::2]+==+"000"|2983631|现在A将是你想要的整数。|2983632|编辑:你似乎完全改变了你的问题，所以它变得更加琐碎：|2983633|a+=+"102030"[1::2]|2983634|您可以将‘切片’分成字符串，切片语法是seqence_type[start:end:step]，如果第一个参数从第一个索引开始(在本例中将得到"123“而不是"000"，所以我们不能这样做)则可以省略第一个参数，如果省略第二步到末尾，如果省略最后一个参数步骤1，则不能这样做。这样，您甚至可以使用切片来复制数组/字符串|style|CODE|2983635|hello+=+"hello"
hello2+=+hello[:]%2B"2"
print(hello2)
#+"hello2"|2983636|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/functions.html#slice^0|0|0|0|B|4|0|0|0|0|0|0|J|S|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|19|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|1A|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|1B|8|@]|9|@$K|1C|L|1D|1|1E]]|A|$]]|$1|M|3|N|5|F|7|1F|8|@]|9|@]|A|$G|H]]|$1|O|3|P|5|6|7|1G|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1H|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|1I|8|@]|9|@]|A|$G|H]]|$1|U|3|V|5|6|7|1J|8|@$K|1K|L|1L|W|X]]|9|@]|A|$]]|$1|Y|3|Z|5|F|7|1M|8|@]|9|@]|A|$G|H]]|$1|10|3|-4|5|6|7|1N|8|@]|9|@]|A|$]]]|11|$12|$5|13|14|15|A|$16|17]]]]

First, you can only slice on certain sequence objects, like lists and strings. You have an integer. You can still get the result you want however.

first take your number and convert it into a string:

<pre><code>a = 102030
str_a str(a)
</code></pre>

then after doing this, you can use normal <a href="https://docs.python.org/3/library/functions.html#slice" rel="nofollow noreferrer">slice syntax</a>, so you can do this with the start, end, step size

<pre><code>a = int(str_a[1::2])
# a == 0
# but str_a[1::2] == "000"
</code></pre>

now a will be the int you want. 

EDIT: you seem to have changed your question entirely so it becomes even more trivial:

<pre><code>a = "102030"[1::2]
</code></pre>

You are able to 'slice' into strings, slicing syntax is <code>seqence_type[start:end:step]</code> you can omit the first parameter if it starts at the first index (in this case you would get "123" instead of "000", so we can't do that) if you omit the second it goes to the end and if you omit the last parameter step size of 1 is used. In this way you can even use slicing to copy arrays/strings via

<pre><code>hello = "hello"
hello2 = hello[:]+"2"
print(hello2)
# "hello2"
</code></pre>

blocks|key|1271297|text|在python中，如果您有一个list、string等(称为a)，您可以通过提供希望a结束的索引范围来访问它的元素。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1271298|例如，如果是a+=+"abcdef"，a[0:6]将返回"abcdef"。也就是说，a[0:6]将对应于“a的前六个元素”。|1271299|如果给它a[1:5]，输出将是bcde，即从1到4的元素(5不包括在内)。|1271300|您还可以要求它通过指定第三个值(如a[0:6:2]+)跳过若干个元素。这意味着，从第0元素开始，在中间跳到第6跳。|1271301|要回答你的问题，你需要string[1:6:2]|1271302|entityMap^0|U|1|16|1|0|6|C|J|6|S|8|16|6|0|4|6|F|4|0|H|8|0|B|D|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|13|8|@$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|18|8|@$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|L|3|M|5|6|7|1B|8|@$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|N|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|O|$]]

In python, if you have a list, string etc(call it <code>a</code>), you can access its elements by providing the range of indexes you want the <code>a</code> over. 

For example, if <code>a = "abcdef"</code>, <code>a[0:6]</code> will give <code>"abcdef"</code> back. That is,<code>a[0:6]</code> would correspond to 'first six elements of a'. 

If you give it <code>a[1:5]</code> the output would be <code>bcde</code>, that is elements from 1 through 4(5 is not inclusive). 

You can also ask it to skip a certian number of elements by specifying a third value such as <code>a[0:6:2]</code>. This would mean, start from the 0th element, go up to 6th skip one in the middle. 

To answer your question, you would need <code>string[1:6:2]</code>

I am wondering how I can extract every other digit (starting with the second) from a number assigned to a variable in python.

For example, I want this:

<pre><code>a= '102030'
</code></pre>

to become

<pre><code>a= '000'
</code></pre>

I hear this is possible to do using "slicing," I'm just not sure how.

Any help is appreciated!

How to select every other using slicing 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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我想知道如何从python中分配给变量的数字中提取每一个其他数字(从第二个数字开始)。例如，我想要这样：a= '102030'成为a= '000'我听说这是可以用“切片”，我只是不确定如何做。任何帮助都是非常感谢的！

问如何使用Python中的切片来选择其他的
EN

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问如何使用Python中的切片来选择其他的EN