entityMap|blocks|key|988v5|text|我认为问题是，你可能有一个字符串之外的/，它给你一个除以年的月份除以天，也许可以检查一下start变量是不是一开始就是一个字符串？因为一旦数学运算完成，将其转换为字符串只会转换结果，而不会转换字符串本身。请看这里：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|an9gp|start+=+8/1986/20
start+=+str(start)
start+=+list(start.split("-"))
print(start)|code-block|syntax|javascript|71bil|当日期没有用引号括起来时，它将其视为等式并返回：|3jdc4|#['0.0002014098690835851']|af3pv|当它在引号中时，它可以工作，但你的拆分是不正确的|esp47|start+=+'8/1986/20'
start+=+str(start)
start+=+list(start.split("-"))
print(start)|daf9h|它返回|67s4c|#['8/1986/20']|74n63|你想要的是|cksdg|start+=+'8/1986/20'
start+=+str(start)
start+=+list(start.split("/"))
print(start)|65jgm|它返回：|cgvs4|#['8',+'1986',+'20']|10luh^0|J|1|0|0|0|0|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|16|9|@$A|17|B|18|C|D]]|E|@]|F|$]]|$2|G|4|H|6|I|8|19|9|@]|E|@]|F|$J|K]]|$2|L|4|M|6|7|8|1A|9|@]|E|@]|F|$]]|$2|N|4|O|6|I|8|1B|9|@]|E|@]|F|$J|K]]|$2|P|4|Q|6|7|8|1C|9|@]|E|@]|F|$]]|$2|R|4|S|6|I|8|1D|9|@]|E|@]|F|$J|K]]|$2|T|4|U|6|7|8|1E|9|@]|E|@]|F|$]]|$2|V|4|W|6|I|8|1F|9|@]|E|@]|F|$J|K]]|$2|X|4|Y|6|7|8|1G|9|@]|E|@]|F|$]]|$2|Z|4|10|6|I|8|1H|9|@]|E|@]|F|$J|K]]|$2|11|4|12|6|7|8|1I|9|@]|E|@]|F|$]]|$2|13|4|14|6|I|8|1J|9|@]|E|@]|F|$J|K]]|$2|15|4|-4|6|7|8|1K|9|@]|E|@]|F|$]]]]

I think the problem is that you possibly have the <code>/</code> outside of a string, which is giving you a the month divided by year divided by day, maybe check to see if the start variable was a string to begin with? as once the math is done converting it to a string will only convert the result, not the string itself.
see here:
<pre><code>start = 8/1986/20
start = str(start)
start = list(start.split(&quot;-&quot;))
print(start)
</code></pre>
When the date is not enclosed in quotations, it treats it as an equation and returns:
<pre><code>#['0.0002014098690835851']
</code></pre>
when it is in quotations it works, but your split is not correct
<pre><code>start = '8/1986/20'
start = str(start)
start = list(start.split(&quot;-&quot;))
print(start)
</code></pre>
it returns
<pre><code>#['8/1986/20']
</code></pre>
what you want is
<pre><code>start = '8/1986/20'
start = str(start)
start = list(start.split(&quot;/&quot;))
print(start)
</code></pre>
which returns:
<pre><code>#['8', '1986', '20']
</code></pre>

entityMap|0|type|LINK|mutability|MUTABLE|data|url|https://stackoverflow.com/questions/30071886/how-to-get-current-time-in-python-and-break-up-into-year-month-day-hour-minu|blocks|key|45nj9|text|+start+=+list(start.split("-"))|code-block|depth|inlineStyleRanges|entityRanges|syntax|javascript|6gi6q|这将查找"-“。如果您的格式包含"/“，请使用|unstyled|bqljr|++++start+=+list(start.split("/"))|djjdk|但正如评论中所给出的，How+to+get+current+time+in+python+and+break+up+into+year,+month,+day,+hour,+minute?更好|offset|length|ccl50^0|0|0|0|B|2B|0|0^^$0|$1|$2|3|4|5|6|$7|8]]]|9|@$A|B|C|D|2|E|F|U|G|@]|H|@]|6|$I|J]]|$A|K|C|L|2|M|F|V|G|@]|H|@]|6|$]]|$A|N|C|O|2|E|F|W|G|@]|H|@]|6|$I|J]]|$A|P|C|Q|2|M|F|X|G|@]|H|@$R|Y|S|Z|A|10]]|6|$]]|$A|T|C|-4|2|M|F|11|G|@]|H|@]|6|$]]]]

<pre><code> start = list(start.split(&quot;-&quot;))
</code></pre>
This looks for &quot;-&quot;. If your format has &quot;/&quot; please use
<pre class="lang-py prettyprint-override"><code> start = list(start.split(&quot;/&quot;))
</code></pre>
But as given in the comment <a href="https://stackoverflow.com/questions/30071886/how-to-get-current-time-in-python-and-break-up-into-year-month-day-hour-minu">How to get current time in python and break up into year, month, day, hour, minute?</a> is better

entityMap|blocks|key|dgcht|text|根据你得到的输出，你的代码必须是这样的-|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3ha7|++++start+=+5-2009-30
++++start+=+str(start)
++++start+=+list(start.split("-"))
++++print(start)|code-block|syntax|javascript|c6lq0|你的代码所做的，就是减去5，2009和30。因为您在该变量中给出的值是整数，而"-“符号表示减法，所以它将数字相除。然后，它将答案转换为字符串，然后转换为列表，并打印差异。这就是为什么你得到了错误的输出。您可以通过将数据放在引号之间来避免这种情况。转换为字符串意味着，它只是将答案转换为字符串，这就是为什么即使您键入str()函数，它也不会给出任何结果|cg9t9|所以你的最终代码应该是-|dpjho|start+=+"5-2009-30"
start+=+list(start.split("-"))
print(start)|83e1j^0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|O|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|P|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|Q|9|@]|A|@]|B|$]]|$2|J|4|K|6|7|8|R|9|@]|A|@]|B|$]]|$2|L|4|M|6|E|8|S|9|@]|A|@]|B|$F|G]]|$2|N|4|-4|6|7|8|T|9|@]|A|@]|B|$]]]]

According to the output you are getting, this is how your code must be -
<pre><code> start = 5-2009-30
 start = str(start)
 start = list(start.split(&quot;-&quot;))
 print(start)
</code></pre>
What your code is doing, is that, it subtracts 5, 2009, and 30. Because the values you gave in that variable are integral, and the &quot;-&quot; symbol means subtraction, it divides the numbers. Then It is converting the answer into a string, then converting into a list, and printing the difference. That is why you are getting the wrong output. You can avoid this by, placing the data between the quotating marks. Converting into string means that, it just converts the answer into a string, thats why even when you typed the str() function, it didnt give any result
So your final code should be -
<pre><code>start = &quot;5-2009-30&quot;
start = list(start.split(&quot;-&quot;))
print(start)
</code></pre>

entityMap|blocks|key|5l0u9|text|这是你所要求的一个简单的答案。确保将日期声明为字符串，这样数字就不会彼此分开！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|b8dnj|start+="5/2009/30"

start+=+list(start.split("/"))

print(start)|code-block|syntax|javascript|6svf8^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|I|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|J|9|@]|A|@]|B|$F|G]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

This is a simple answer to what you are asking for. Make sure to state the date as a string so that the numbers would not divide each other!
<pre><code>start =&quot;5/2009/30&quot;

start = list(start.split(&quot;/&quot;))

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

I want to convert a date to a list format and then access its elements. The current code I am using is -
<pre><code> start = str(start)
 start = list(start.split(&quot;-&quot;))
 print(start)
</code></pre>
where start will be a date in the format - MM/YYYY/DD
But the output for this is -
<pre><code>['0.0019821605550049554']
</code></pre>
I don't understand what am I doing wrong

converting a date to list 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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 我希望将日期转换为列表格式，然后访问其元素。我当前使用的代码是-  start = str(start) start = list(start.split("-")) print(start) 其中开始日期的格式为- MM/YYYY/DD 但是它的输出是- ['0.0019821605550049554'] 我不明白...

问将日期转换为python中的列表
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