Find the indices of elements greater than x

Question

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!

Asked By: Oliver Amundsen

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Source

Answer 1

>>> [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]

Answered By: Aesthete

Answer 2

>>> 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])

Answered By: Joran Beasley

Answer 3

using filter built-in function is fine

>>>a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
>>>filter(lambda x : x < 4, a)
[1, 2, 3]

Explanation

filter(FUN, Iterable)

this expression will iterate all element from Iterable and supply to FUN function as argument, if return is True ,then the arugment will be append to a internal list

lambda x: x > 4

this means a anonymous function that will take a argument and test it if bigger than 4, and return True of False value

Your solution

if you are try to delete all elements larger than 4 ,then try blow

>>> a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> filter(lambda x: x<4 ,a)
[1, 2, 3]

Answered By: Shawn Zhang

Answer 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 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]

Answered By: Shawn Zhang

Answer 5

The simplest in my eyes would be to use numpy

X[np.array(a)>4]#X needs to be np.array as well

Explanation:
np.array converts a to an array.

np.array(a)>4 gives a bool array with all the elements that should be kept

And X is filtered by the bool array so only the elements where a is greater than 4 are selected (and the rest discarded)

Answered By: Okapi575

Answer 6

I guess I came here a bit late (while things got easier using Numpy)..

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]

Answered By: singrium

Answer 7

looping is slow, using divide and conquer method. code in C++

// find index whose value is equal to or greater than "key" in an ordered vector.
// note: index may be equal to indices.size()
size_t StartIndex(const std::vector<int>& indices, int key)
{
    if (indices.empty() || key <= indices[0])
        return 0;

    if (key > indices.back())
        return indices.size();

    size_t st = 0;
    size_t end = indices.size() - 1;

    while (true)
    {
        if ((end - st) < 2)
            return (indices[st] < key) ? end : st;

        size_t mid = ((st + end) >> 1);  // (st + end) / 2

        if (indices[mid] == key)
            return mid;

        (indices[mid] < key ? st : end) = mid;
    }
}

Answered By: JSong

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