Given the following vector,
a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
I need to identify the indices of "a" whose elements are >= than 4, like this:
idx = [3, 4, 5, 6, 7, 8]
The info in "idx" will be used to delete the elements from another list X (X has the same number of elements that "a"):
del X[idx] #idx is used to delete these elements in X. But so far isn't working.
I heard that numpy might help. Any ideas? Thanks!
>>> [i for i,v in enumerate(a) if v > 4]
[4, 5, 6, 7, 8]
enumerate返回数组中每个项目的索引和值。因此，如果该值v大于4，则将索引包含i在新数组中。
或者您可以就地修改您的列表并排除上述所有值4。
>>> a[:] = [x for x in a if x<=4]
>>> a
[1, 2, 3, 4]
好的，我明白你的意思，一行 Python 就足够了：
使用列表理解
[ j for (i,j) in zip(a,x) if i >= 4 ]
# a will be the list compare to 4
# x another list with same length
Explanation:
>>> a
[1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> x
['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'j']
Zip 函数将返回一个元组列表
>>> zip(a,x)
[(1, 'a'), (2, 'b'), (3, 'c'), (4, 'd'), (5, 'e'), (6, 'f'), (7, 'g'), (8, 'h'), (9, 'j')]
列表推导是在“in”之后的列表中循环元素的快捷方式，并使用表达式评估元素，然后将结果返回到列表中，您也可以添加要返回的结果的条件
>>> [expression(element) for **element** in **list** if condition ]
此代码只返回所有压缩的对。
>>> [(i,j) for (i,j) in zip(a,x)]
[(1, 'a'), (2, 'b'), (3, 'c'), (4, 'd'), (5, 'e'), (6, 'f'), (7, 'g'), (8, 'h'), (9, 'j')]
我们所做的是通过指定“if”后跟一个布尔表达式来添加一个条件
>>> [(i,j) for (i,j) in zip(a,x) if i >= 4]
[(4, 'd'), (5, 'e'), (6, 'f'), (7, 'g'), (8, 'h'), (9, 'j')]
使用 Itertools
>>> [ _ for _ in itertools.compress(d, map(lambda x: x>=4,a)) ]
# a will be the list compare to 4
# d another list with same length
在 Python 中使用 itertools.compress单行来完成关闭此任务
>>> a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> d = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'j'] # another list with same length
>>> map(lambda x: x>=4, a) # this will return a boolean list
[False, False, False, True, True, True, True, True, True]
>>> import itertools
>>> itertools.compress(d, map(lambda x: x>4, a)) # magic here !
<itertools.compress object at 0xa1a764c> # compress will match pair from list a and the boolean list, if item in boolean list is true, then item in list a will be remain ,else will be dropped
#below single line is enough to solve your problem
>>> [ _ for _ in itertools.compress(d, map(lambda x: x>=4,a)) ] # iterate the result.
['d', 'e', 'f', 'g', 'h', 'j']
对itertools.compress的解释，我想这对你的理解会很清楚：
>>> [ _ for _ in itertools.compress([1,2,3,4,5],[False,True,True,False,True]) ]
[2, 3, 5]
>>> import numpy as np
>>> a = np.array(range(1,10))
>>> indices = [i for i,v in enumerate(a >= 4) if v]
>>> indices
[3, 4, 5, 6, 7, 8]
>>> mask = a >= 4
>>> mask
array([False, False, False, True, True, True, True, True, True], dtype=boo
l)
>>> a[mask]
array([4, 5, 6, 7, 8, 9])
>>> np.setdiff1d(a,a[mask])
array([1, 2, 3])
在我看来最简单的就是使用 numpy
X[np.array(a)>4]#X needs to be np.array as well
说明： np.array 将 a 转换为数组。
np.array(a)>4 给出一个 bool 数组，其中包含应保留的所有元素
并且 X 被 bool 数组过滤，因此只选择 a 大于 4 的元素（其余的被丢弃）
我想我来这里有点晚了（虽然使用 Numpy 变得更容易了）..
import numpy as np
# Create your array
a = np.arange(1, 10)
# a = array([1, 2, 3, 4, 5, 6, 7, 8, 9])
# Get the indexes/indices of elements greater than 4
idx = np.where(a > 4)[0]
# idx = array([4, 5, 6, 7, 8])
# Get the elements of the array that are greater than 4
elts = a[a > 4]
# elts = array([5, 6, 7, 8, 9])
# Convert idx(or elts) to a list
idx = list(idx)
#idx = [4, 5, 6, 7, 8]
使用过滤器内置功能很好
>>>a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
>>>filter(lambda x : x < 4, a)
[1, 2, 3]
解释
过滤器（有趣，可迭代）
此表达式将迭代 Iterable 中的所有元素并提供给 FUN 函数作为参数，如果 return 为 True ，则参数将附加到内部列表
λx：x > 4
这意味着一个匿名函数，它将接受一个参数并测试它是否大于 4，并返回 True of False 值
您的解决方案
如果您尝试删除所有大于 4 的元素，然后尝试打击
>>> a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> filter(lambda x: x<4 ,a)
[1, 2, 3]