blocks|key|212796|text|filter(lambda+n:+isinstance(n,+int),+[1,2,"three"])|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|212797|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>filter(lambda n: isinstance(n, int), [1,2,"three"])
</code></pre>

blocks|key|4514613|text|列表理解。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4514614|list_2+=+[num+for+num+in+list_1+if+isinstance(num,+(int,float))]|code-block|syntax|javascript|4514615|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|$]]

List comprehensions.

<pre><code>list_2 = [num for num in list_1 if isinstance(num, (int,float))]
</code></pre>

blocks|key|1514312|text|list_2+=+[i+for+i+in+list_1+if+isinstance(i,+(int,+float))]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1514313|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>list_2 = [i for i in list_1 if isinstance(i, (int, float))]
</code></pre>

blocks|key|1298096|text|list_2+=+[i+for+i+in+list_1+if+isinstance(i,+(int,+float))]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1298097|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|$]]

blocks|key|212809|text|>>>+[+i+for+i+in+list_1+if+not+str(i).replace("+","").isalpha()+]
[1,+123131.13099999999,+9.9999999999999995e-07,+34.124512352650001]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|212810|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>&gt;&gt;&gt; [ i for i in list_1 if not str(i).replace(" ","").isalpha() ]
[1, 123131.13099999999, 9.9999999999999995e-07, 34.124512352650001]
</code></pre>

blocks|key|212718|text|这应该是Python+2中最高效和最短的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|212719|import+operator
filter(operator.isNumberType,+list_1)|code-block|syntax|javascript|212720|在Python+3中是这样的：|212721|import+numbers
[x+for+x+in+list_1+if+isinstance(x,+numbers.Number)]|212722|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|$]]

This should be the most efficient and shortest in Python 2:
<pre><code>import operator
filter(operator.isNumberType, list_1)
</code></pre>
This in Python 3:
<pre><code>import numbers
[x for x in list_1 if isinstance(x, numbers.Number)]
</code></pre>

blocks|key|1298172|text|SilentGhost方式的缩写|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1298173|list_2+=+[i+for+i+in+list_1+if+isinstance(i,+(int,+float))]+|code-block|syntax|javascript|1298174|至|1298175|list_2+=+[i+for+i+in+list_1+if+not+isinstance(i,+str)]|1298176|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|$]]

for short of SilentGhost way

<pre><code>list_2 = [i for i in list_1 if isinstance(i, (int, float))] 
</code></pre>

to

<pre><code>list_2 = [i for i in list_1 if not isinstance(i, str)]
</code></pre>

blocks|key|212842|text|我认为最简单的方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|212843|[s+for+s+in+myList+if+s.isdigit()]|code-block|syntax|javascript|212844|希望它能帮上忙！|212845|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I think the easiest way is:

<pre><code>[s for s in myList if s.isdigit()]
</code></pre>

Hope it helps!

blocks|key|1512725|text|仅当列表中的数字已转换为适当的类型(int、float)时，建议的所有解决方案才有效。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1512726|我发现自己有一个来自sqlite3的fetchall函数的列表。那里的所有元素都被格式化为str，即使其中一些元素实际上是整数。|1512727|cur.execute('SELECT+column1+FROM+table1+WHERE+column2+=+?',+(some_condition,+))
list_of_tuples+=+cur.fetchall()|code-block|syntax|javascript|1512728|问题中的等价物应该是这样的：|1512729|list_1+=+[+'asdada',+'1',+'123131',+'blaa+adaraerada',+'0',+'34',+'stackoverflow+is+awesome'+]|1512730|对于这种情况，为了得到一个只包含整数的列表，这是我找到的另一种选择：|1512731|list_of_numbers+=+[]
for+tup+in+list_of_tuples:
++++try:
++++++++list_of_numbers.append(int(tup[0]))
++++except+ValueError:
++++++++pass|1512732|list_of_numbers将只包含初始列表中的所有整数。|BOLD|1512733|entityMap^0|I|3|M|5|0|A|7|I|8|19|3|0|0|0|0|0|0|0|F|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|14|8|@$9|15|A|16|B|C]|$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|1B|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1C|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|1D|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1E|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|1F|8|@]|D|@]|E|$K|L]]|$1|U|3|V|5|6|7|1G|8|@$9|1H|A|1I|B|W]]|D|@]|E|$]]|$1|X|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]]|Y|$]]

All solutions proposed work only if the numbers inside the list are already converted to the appropriate type (<code>int</code>, <code>float</code>).

I found myself with a list coming from the <code>fetchall</code> function of <code>sqlite3</code>. All elements there were formated as <code>str</code>, even if some of those elements were actually integers.

<pre><code>cur.execute('SELECT column1 FROM table1 WHERE column2 = ?', (some_condition, ))
list_of_tuples = cur.fetchall()
</code></pre>

The equivalent from the question would be having something like:

<pre><code>list_1 = [ 'asdada', '1', '123131', 'blaa adaraerada', '0', '34', 'stackoverflow is awesome' ]
</code></pre>

For such a case, in order to get a list with the integers only, this is the alternative I found:

<pre><code>list_of_numbers = []
for tup in list_of_tuples:
 try:
 list_of_numbers.append(int(tup[0]))
 except ValueError:
 pass
</code></pre>

list_of_numbers will contain only all integers from the initial list.

blocks|key|4514792|text|list_1+=+[+'asdada',+1,+123131.131,+'blaa+adaraerada',+0.000001,+34.12451235265,+'stackoverflow+is+awesome'+]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4514793|第一个示例：|unstyled|4514794|list_2+=+[x+for+x+in+list_1+if+type(x)+==+int+or+type(x)+==+float+]
print(list_2)|4514795|第二个例子：|4514796|list_2+=+[x+for+x+in+list_1+if+isinstance(x,+(int,+float))+]
print(list_2)|4514797|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|P|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|M|3|-4|5|F|7|T|8|@]|9|@]|A|$]]]|N|$]]

<pre><code>list_1 = [ 'asdada', 1, 123131.131, 'blaa adaraerada', 0.000001, 34.12451235265, 'stackoverflow is awesome' ]
</code></pre>
first example:
<pre><code>list_2 = [x for x in list_1 if type(x) == int or type(x) == float ]
print(list_2)
</code></pre>
second example:
<pre><code>list_2 = [x for x in list_1 if isinstance(x, (int, float)) ]
print(list_2)
</code></pre>

Simple Question:

<pre><code>list_1 = [ 'asdada', 1, 123131.131, 'blaa adaraerada', 0.000001, 34.12451235265, 'stackoverflow is awesome' ]
</code></pre>

I want to create a <code>list_2</code> such that it only contains the numbers:

<pre><code>list_2 = [ 1, 123131.131, 0.000001, 34.12451235265 ]
</code></pre>

Is there simplistic way of doing this, or do I have to resort to checking the variable type of each list item and only output the numerical ones?

Is there a way to output the numbers only from a python list?

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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简单的问题：list_1 = [ 'asdada', 1, 123131.131, 'blaa adaraerada', 0.000001, 34.12451235265, 'stackoverflow is awesome' ]我想创建一个list_2，让它只包含以下数字：list_2 = [ 1, 123131.1...

问有没有办法只输出python列表中的数字？
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问有没有办法只输出python列表中的数字？EN