Find the indices of elements greater than x

>>> [i for i,v in enumerate(a) if v > 4][4, 5, 6, 7, 8]

enumerate returns the index and value of each item in an array. So if the value v is greater than 4, include the index i in the new array.

Or you can just modify your list in place and exclude all values above 4.

>>> a[:] = [x for x in a if x<=4]>>> a [1, 2, 3, 4]

OK, I understand what you mean and a Single line of Python will be enough:

using list comprehension

[ j for (i,j) in zip(a,x) if i >= 4 ]# a will be the list compare to 4# x another list with same lengthExplanation:>>> a[1, 2, 3, 4, 5, 6, 7, 8, 9]>>> x['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'j']

Zip function will return a list of tuples

>>> zip(a,x)[(1, 'a'), (2, 'b'), (3, 'c'), (4, 'd'), (5, 'e'), (6, 'f'), (7, 'g'), (8, 'h'), (9, 'j')]

List comprehension is a shortcut to loop an element over list which after "in", and evaluate the element with expression, then return the result to a list, also you can add condition on which result you want to return

>>> [expression(element) for **element** in **list** if condition ]

This code does nothing but return all pairs that zipped up.

>>> [(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')]

What we do is to add a condition on it by specify "if" follow by a boolean expression

>>> [(i,j) for (i,j) in zip(a,x) if i >= 4][(4, 'd'), (5, 'e'), (6, 'f'), (7, 'g'), (8, 'h'), (9, 'j')]

using 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

Use itertools.compress with single line in Python to finish close this task

>>> 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']

Explanation for itertools.compress, I think this will be clear for your understanding:

>>> [ _ 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>>> maskarray([False, False, False,  True,  True,  True,  True,  True,  True], dtype=bool)>>> a[mask]array([4, 5, 6, 7, 8, 9])>>> np.setdiff1d(a,a[mask])array([1, 2, 3])