Finding the index of an element in nested lists in python

Question

I am trying to get the index of an element in nested lists in python - for example [[a, b, c], [d, e, f], [g,h]] (not all lists are the same size). I have tried using

strand_value= [x[0] for x in np.where(min_value_of_non_empty_strands=="a")]

but this is only returning an empty list, even though the element is present. Any idea what I'm doing wrong?

In this instance, 0, I would then try using x[1] to get 0 again - I want to know which list it's in and which position within the list — biophys_chem, Nov 26 '15 at 12:24
I am trying to find the index of any given string, for which I do not know its position or which list it is in - if I can get 0 0 straight away, I would find that very helpful — biophys_chem, Nov 26 '15 at 12:27
If you have a list of strings, then it is much easier to solve than a list of lists of stings. — DainDwarf, Nov 26 '15 at 12:31
Why the `numpy` tag? Your expression using `np.where` does not make sense. — hpaulj, Nov 26 '15 at 15:44
Does this answer your question? [Find the index of an item in a list of lists](https://stackoverflow.com/questions/9553638/find-the-index-of-an-item-in-a-list-of-lists) — Georgy, Sep 01 '20 at 17:58

score 6 · Accepted Answer · answered Nov 26 '15 at 12:29

6

def find_in_list_of_list(mylist, char):
    for sub_list in mylist:
        if char in sub_list:
            return (mylist.index(sub_list), sub_list.index(char))
    raise ValueError("'{char}' is not in list".format(char = char))

example_list = [['a', 'b', 'c'], ['d', 'e', 'f'], ['g', 'h']]

find_in_list_of_list(example_list, 'b')
(0, 1)

answered Nov 26 '15 at 12:29

DainDwarf

1,593
9
19

1

This is O(N^2) and will be slow if `mylist` is long. – chthonicdaemon Nov 26 '15 at 12:36

score 2 · Answer 2 · answered Nov 26 '15 at 12:32

suppose your list is like this:

lst = [['a', 'b', 'c'], ['d', 'e', 'f'], ['g','h']]
list_no = 0
pos = 0
for x in range(0,len(lst)):
    try:
        pos = lst[x].index('e')
        break
    except:
        pass

list_no = x

list_no gives the list number and pos gives the position in that list

The6thSense · Answer 3 · 2015-11-26T12:39:37.617

You could do this using List comprehension and enumerate

Code:

lst=[["a", "b", "c"], ["d", "e", "f"], ["g","h"]]
check="a"
print ["{} {}".format(index1,index2) for index1,value1 in enumerate(lst) for index2,value2 in enumerate(value1) if value2==check]

Output:

['0 0']

Steps:

I have enumerated through the List of List and got it's index and list
Then I have enumerated over the gotten list and checked if it matches the check variable and written it to list if so

This gives all possible output

i.e.)

Code2:

lst=[["a", "b", "c","a"], ["d", "e", "f"], ["g","h"]]
check="a"
print ["{} {}".format(index1,index2) for index1,value1 in enumerate(lst) for index2,value2 in enumerate(value1) if value2==check]

Gives:

['0 0', '0 3']

Notes:

You can easily turn this into list of list instead of string if you want

score 0 · Answer 4 · answered Nov 26 '15 at 12:33

does this suffice?

array = [['a', 'b', 'c'], ['d', 'e', 'f'], ['g', 'h']]
for subarray in array:
    if 'a' in subarray:
        print(array.index(subarray), '-', subarray.index('a'))

This will return 0 - 0. First zero is the index of the subarray inside array, and the last zero is the 'a' index inside subarray.

Yerbol Sapar · Answer 5 · 2022-05-04T14:36:26.480

Reworked Bhrigu Srivastava's proposal:

def findinlst(lst, val):
    for x in range(0, len(lst)):
        try:
            pos = lst[x].index(val)
            return [x, pos]
        except:
            continue
    return [False, False]  # whatever one wants to get if value not found

arr = [['a', 'b', 'c'], ['d', 'e', 'f'], ['g', 'h']]

findinlst(arr, 'b')
[0, 1]

findInLst(arr, 'g')
[2, 0]

findinlst(arr, 'w')
[False, False]

Finding the index of an element in nested lists in python

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