[英]Remove elements from python list based on filter
I have 3 lists in Python. 我在Python中有3个列表。
list_a = [10., 20., 30., 12.]
list_b = [30., 20., 60., 12.]
list_c = [10., 80., 90., 12.]
I want to drop those elements in list_b
and list_c
where values in list_a
are <= 15
. 我想删除那些元素在
list_b
和list_c
其中值list_a
是<= 15
。 Therefore results become: 因此结果变为:
list_b = [20., 60.]
list_c = [80., 90.]
Is there a way to do this without for loops? 有没有办法在没有循环的情况下做到这一点? (list comprehensions are ok)
(列表理解还可以)
If you are using numpy as you said you are, in comments, you can simply create a boolean index from a
and mask elements in b
and c
: 如果您正在使用numpy的,你说你在评论,你可以简单地创建从布尔指数
a
中和掩码元素b
和c
:
import numpy as np
list_a = np.array([10., 20., 30., 12.])
list_b = np.array([30., 20., 60., 12.])
list_c = np.array([10., 80., 90., 12.])
list_b = list_b[list_a>15]
list_c = list_c[list_a>15]
print list_a
print list_b
print list_c
Output: 输出:
[ 10. 20. 30. 12.]
[ 20. 60.]
[ 80. 90.]
You can convert the lists_b
and list_c
to Python list type
by .tolist()
method. 您可以通过
.tolist()
方法将lists_b
和list_c
转换为Python list type
。
You can use the little known itertools.compress
class to do this. 您可以使用鲜为人知的
itertools.compress
类来执行此操作。 See Filter list using Boolean index arrays 请参阅使用布尔索引数组筛选列表
>>> import itertools
>>> list_a = [10., 20., 30., 12.]
>>> list_b = [30., 20., 60., 12.]
>>> list_c = [10., 80., 90., 12.]
>>> bool_list = [item > 15 for item in list_a]
>>> bool_list
[False, True, True, False]
>>> list_b = list(itertools.compress(list_b, bool_list))
>>> list_b
[20.0, 60.0]
>>> list_c = list(itertools.compress(list_c, bool_list))
>>> list_c
[80.0, 90.0]
Can you use list comprehensions? 你能使用列表推导吗?
You could do it like this: 你可以这样做:
list_a = [10., 20., 30., 12.]
list_b = [30., 20., 60., 12.]
list_c = [10., 80., 90., 12.]
list_b = [ el for (i, el) in enumerate(list_b) if (list_a[i] > 15) ]
list_c = [ el for (i, el) in enumerate(list_c) if (list_a[i] > 15) ]
Snipper was written here, I haven't tested it, but you see the general idea. Snipper写在这里,我还没有测试过,但你看到了一般的想法。
I assumed that all lists are the same length. 我假设所有列表都是相同的长度。 If
list_a
is shorter and you want to drop elements that are on the missing positions, you can do it like this: 如果
list_a
较短并且您想要删除缺失位置上的元素,则可以这样做:
list_b = [ el for (i, el) in enumerate(list_b) if (i<len(list_a) and list_a[i] > 15) ]
and if you want to keep them, just reverse the sign and boolean operator: 如果你想保留它们,只需反转符号和布尔运算符:
list_b = [ el for (i, el) in enumerate(list_b) if (i>=len(list_a) or list_a[i] > 15) ]
list_a = [10., 20., 30., 12.]
list_b = [30., 20., 60., 12.]
list_c = [10., 80., 90., 12.]
drop = [i for i,v in enumerate(list_a) if v <=15]
b = [v for i,v in enumerate(list_b) if not i in drop]
c = [v for i,v in enumerate(list_c) if not i in drop]
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.