Getting index with list comprehension in a list in Python

Question

Here are two fonctions which (I thought) should do the same thing but actually do not.

It seems that with the list comprehension, the index taken is the first that could correspond, so when you have the same value at different index, there is an ambiguity.

Is there a way to modify the list comprehension in filter2 so get the same result as in filter1 ?

  L = [98.75011926342906,
 97.8178200008178,
 98.6138182016438,
 98.55520874507613,
 98.25262038791283,
 98.75011926342906,
 99.06770073738875,
 98.66970163697574,
 98.56611283001895,
 98.47751713985852, 
 98.66970163697574,
 97.8178200008178]


def filter1(L, threshold=98.7):
    items = []
    for i in range(len(L)):
        if L[i] < threshold:
            items.append(i)
    return items

def filter2(L, threshold=98.7):
    items = [L.index(x) for x in L if  x <= threshold]
    return items

print filter1(L)
>>> [1, 2, 3, 4, 7, 8, 9, 10, 11]
print filter2(L)
>>> [1, 2, 3, 4, 7, 8, 9, 7, 1]

pay attention in the future to the small bits. What I mean is that in filter1 you use < sign, in filter2 you use <=. The result is also different because these two filters use different logic, index returns the value of the first index found (that is where the flaw is). — Damian Chrzanowski
– Damian Chrzanowski, Commented Dec 6, 2016 at 19:37

Natha · Accepted Answer · 2016-12-06 21:59:12Z

15

You can use enumerate as a helper here:

bad_items = [i for i, x in enumerate(L) if x <= threshold]

enumerate will give you pairs of (index, value) which you can unpack in comprehension
(into i, x). Then you only take i if x <= threshold.

edited Dec 6, 2016 at 21:59

Natha

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answered Dec 6, 2016 at 19:24

mgilson

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3 Comments

Moinuddin Quadri Over a year ago

I think you mean for i, x in enumerate(L)

brosenheim Over a year ago

Does enumerate generate indices, or simply count iterations? It is often written as generating indices, but if you apply this code to a dataframe column with indices that do not start at 0 or are non-sequential (for instance, a slice of a dataframe), then this will not return those indices. Rather a sequence of integers beginning at 0 is returned.

mgilson Over a year ago

enumerate just yields pairs. The first value of the pair is an incrementing integer, the second value is pulled from the iterable passed to enumerate.

coredump · Accepted Answer · 2016-12-06 19:29:13Z

3

The reason you have index 7 instead of 10 is because you have duplicate elements and index returns the smallest index at which the value is present. Besides, searching for the index takes linear time too. Your whole loop is being quadratic.

answered Dec 6, 2016 at 19:29

coredump

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Comments

Maurice Meyer · Accepted Answer · 2016-12-06 19:29:39Z

0

You can use enumerate, which assigns the position of the loop into i, x is assigned with the current value.

def filter2(L, threshold=98.7):
    items = [i for (i, x) in enumerate(L) if x <= 98.7]
    return items

answered Dec 6, 2016 at 19:29

Maurice Meyer

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