entityMap|blocks|key|9if7u|text|您可以获取要跳过的号码列表。然后将每个列表项与range(num%2B1)中的列表项进行比较。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|6651o|def+count_down_skip(start,+skip+=+[]):
++++return+[num+for+num+in+reversed(range(start+%2B+1))+if+num+not+in+skip]

print(count_down_skip(10,[1,4,3]))
#[10,+9,+8,+7,+6,+5,+2,+0]|code-block|syntax|javascript|b5d07^0|N|C|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|M|9|@$A|N|B|O|C|D]]|E|@]|F|$]]|$2|G|4|H|6|I|8|P|9|@]|E|@]|F|$J|K]]|$2|L|4|-4|6|7|8|Q|9|@]|E|@]|F|$]]]]

You can take the list of numbers to be skipped. You then compare each list item with the one in <code>range(num+1)</code>.
<pre class="lang-py prettyprint-override"><code>def count_down_skip(start, skip = []):
 return [num for num in reversed(range(start + 1)) if num not in skip]

print(count_down_skip(10,[1,4,3]))
#[10, 9, 8, 7, 6, 5, 2, 0]
</code></pre>

entityMap|blocks|key|dj7im|text|count_down_skip函数中的参数太多。如果想要输入多个数字，可以使用列表，如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|9df94|def+count_down_skip(start,+skip=0):
++++"""
++++Counting+down+a+sequence+with+a+skip+value,
++++from+a+defined+start+point+in+reversed+order.

++++Args:
++++++++start:+start+loop+index.
++++++++skip:+number+to+skip+over.

++++Returns:
++++++++(list):+skipped+list.

++++"""
++++return+[num+for+num+in+reversed(range(start+%2B+1))+if+num+not+in+skip]


print("...+".join(map(str,+count_down_skip(10,+[1,4,3])))+%2B+"!")|code-block|syntax|javascript|2ohnf^0|0|F|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|M|9|@$A|N|B|O|C|D]]|E|@]|F|$]]|$2|G|4|H|6|I|8|P|9|@]|E|@]|F|$J|K]]|$2|L|4|-4|6|7|8|Q|9|@]|E|@]|F|$]]]]

You have too many arguments in the <code>count_down_skip</code> function. You could use a list if you want to input multiple numbers, like this:
<pre><code>def count_down_skip(start, skip=0):
 &quot;&quot;&quot;
 Counting down a sequence with a skip value,
 from a defined start point in reversed order.

 Args:
 start: start loop index.
 skip: number to skip over.

 Returns:
 (list): skipped list.

 &quot;&quot;&quot;
 return [num for num in reversed(range(start + 1)) if num not in skip]


print(&quot;... &quot;.join(map(str, count_down_skip(10, [1,4,3]))) + &quot;!&quot;)
</code></pre>

entityMap|blocks|key|8ot9|text|def+count_down_skip(start,+*args):
++++return+[num+for+num+in+reversed(range(start+%2B+1))+if+num+not+in+args]


print("...+".join(map(str,+count_down_skip(10,1,4,3)))+%2B+"!")|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1vji2|输入的参数不能多于函数定义中的参数。但是，您可以使用*args作为参数。这允许向函数输入可变数量的参数。|unstyled|9te73^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|I|9|@]|A|@]|B|$C|D]]|$2|E|4|F|6|G|8|J|9|@]|A|@]|B|$]]|$2|H|4|-4|6|G|8|K|9|@]|A|@]|B|$]]]]

<pre><code>def count_down_skip(start, *args):
 return [num for num in reversed(range(start + 1)) if num not in args]


print(&quot;... &quot;.join(map(str, count_down_skip(10,1,4,3))) + &quot;!&quot;)
</code></pre>
You cannot input more arguments than in the function definition. You can however use *args as a parameter. This allows to input variable number of parameters to the function.

entityMap|0|type|LINK|mutability|MUTABLE|data|url|https://stackoverflow.com/a/919684/4518341|blocks|key|7c6c4|text|您可以使用*args，并检查num是否为成员：|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|557ca|def+count_down_skip(start,+*skips):
++++return+[num+for+num+in+reversed(range(start%2B1))+if+num+not+in+skips]

print(count_down_skip(10,+1,+4,+3))|code-block|syntax|javascript|be3pt|输出：|2ifan|[10,+9,+8,+7,+6,+5,+2,+0]|6avru^0|5|5|E|3|5|5|0|0|0|0|0^^$0|$1|$2|3|4|5|6|$7|8]]]|9|@$A|B|C|D|2|E|F|W|G|@$H|X|I|Y|J|K]|$H|Z|I|10|J|K]]|L|@$H|11|I|12|A|13]]|6|$]]|$A|M|C|N|2|O|F|14|G|@]|L|@]|6|$P|Q]]|$A|R|C|S|2|E|F|15|G|@]|L|@]|6|$]]|$A|T|C|U|2|O|F|16|G|@]|L|@]|6|$P|Q]]|$A|V|C|-4|2|E|F|17|G|@]|L|@]|6|$]]]]

You can use <a href="https://stackoverflow.com/a/919684/4518341"><code>*args</code></a>, and check if <code>num</code> is a member:
<pre><code>def count_down_skip(start, *skips):
 return [num for num in reversed(range(start+1)) if num not in skips]

print(count_down_skip(10, 1, 4, 3))
</code></pre>
Output:
<pre class="lang-none prettyprint-override"><code>[10, 9, 8, 7, 6, 5, 2, 0]
</code></pre>

entityMap|blocks|key|bq0n1|text|最好的解决方案是，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|fqurg|def+count_down_skip(start,+*skip):
++++return+[num+for+num+in+reversed(range(start+%2B+1))+if+num+not+in+skip]


print("...+".join(map(str,+count_down_skip(10,1,4,3)))+%2B+"!")|code-block|syntax|javascript|ch0ko^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]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

Best Solution is,
<pre><code>def count_down_skip(start, *skip):
 return [num for num in reversed(range(start + 1)) if num not in skip]


print(&quot;... &quot;.join(map(str, count_down_skip(10,1,4,3))) + &quot;!&quot;)
</code></pre>

<pre><code>def count_down_skip(start, skip=0):
 &quot;&quot;&quot;
 Counting down a sequence with a skip value,
 from a defined start point in reversed order.

 Args:
 start: start loop index.
 skip: number to skip over.

 Returns:
 (list): skipped list.

 &quot;&quot;&quot;
 return [num for num in reversed(range(start + 1)) if num != skip]


print(&quot;... &quot;.join(map(str, count_down_skip(10,1))) + &quot;!&quot;)
</code></pre>
this code can output 10 to 0 and without 1
while if 10 to 0 without 1 4 3 (skip these numbers), then how can I do?
I tried change the print :
<pre><code>print(&quot;... &quot;.join(map(str, count_down_skip(10,1,4,3))) + &quot;!&quot;)
</code></pre>
but error happened...
define a function for countdown numbers in python

Define a function for countdown skip numbers

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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 def count_down_skip(start, skip=0):    """    Counting down a sequence with a skip value,    from a defined start point in reversed order.    Args:        star...

问定义一个用于倒计时跳跃数字的函数
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问定义一个用于倒计时跳跃数字的函数EN