有没有办法只从 python 列表中输出数字？
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Is there a way to output the numbers only from a python list?
提问by dassouki
Simple Question:
简单的问题：
list_1 = [ 'asdada', 1, 123131.131, 'blaa adaraerada', 0.000001, 34.12451235265, 'stackoverflow is awesome' ]
I want to create a list_2such that it only contains the numbers:
我想创建一个list_2只包含数字的这样的：
list_2 = [ 1, 123131.131, 0.000001, 34.12451235265 ]
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?
有没有简单的方法可以做到这一点，还是我必须求助于检查每个列表项的变量类型并只输出数字类型？
回答by dalloliogm
This should be the most efficent and shortest:
这应该是最有效和最短的：
import operator
filter(operator.isNumberType, list_1)
Edit: this in python 3000:
编辑：这在python 3000中：
import numbers
[x for x in list_1 if isinstance(x, numbers.Number)]
回答by Ants Aasma
List comprehensions.
列出理解。
list_2 = [num for num in list_1 if isinstance(num, (int,float))]
回答by SilentGhost
list_2 = [i for i in list_1 if isinstance(i, (int, float))]
回答by J0ANMM
All solutions proposed work only if the numbers inside the list are already converted to the appropriate type (int, float).
仅当列表中的数字已转换为适当的类型 ( int, float) 时，提出的所有解决方案才有效。
I found myself with a list coming from the fetchallfunction of sqlite3. All elements there were formated as str, even if some of those elements were actually integers.
我发现自己从未来的一个列表fetchall的功能sqlite3。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。
cur.execute('SELECT column1 FROM table1 WHERE column2 = ?', (some_condition, ))
list_of_tuples = cur.fetchall()
The equivalent from the question would be having something like:
问题中的等价物将具有以下内容：
list_1 = [ 'asdada', '1', '123131', 'blaa adaraerada', '0', '34', 'stackoverflow is awesome' ]
For such a case, in order to get a list with the integers only, this is the alternative I found:
对于这种情况，为了获得仅包含整数的列表，这是我发现的替代方法：
list_of_numbers = []
for tup in list_of_tuples:
try:
list_of_numbers.append(int(tup[0]))
except ValueError:
pass
list_of_numberswill contain only all integers from the initial list.
list_of_numbers将只包含初始列表中的所有整数。
回答by Pontios
I think the easiest way is:
我认为最简单的方法是：
[s for s in myList if s.isdigit()]
Hope it helps!
希望能帮助到你！
回答by Jeff Ober
filter(lambda n: isinstance(n, int), [1,2,"three"])
回答by sykora
list_2 = [i for i in list_1 if isinstance(i, (int, float))]
回答by ghostdog74
>>> [ i for i in list_1 if not str(i).replace(" ","").isalpha() ]
[1, 123131.13099999999, 9.9999999999999995e-07, 34.124512352650001]
回答by wearetherock
for short of SilentGhost way
简称 SilentGhost 方式
list_2 = [i for i in list_1 if isinstance(i, (int, float))]
to
到
list_2 = [i for i in list_1 if not isinstance(i, str)]
