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!

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]

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])

Answer 3

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)

Answer 4

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]

Answer 5

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]

Answer 6

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;
    }
}