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!
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 length
Explanation:
>>> 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]