CODEWITHSUNDEEP
pythonpython-3.xlist
kesoh
kesoh

Reputation: 1

Searching index of element in list of list

I've a list of list and I need to get the index of that element in it.

Suppose I've a list [[0,0,0,'P'],[0,0,0,0],[0,'Q',0,0],[0,0,0,0]].

How can I get index of 'P' and 'Q', if there's only one 'P' and one 'Q'?

I tried: list.index('P'),

but got ValueError: 'P' is not in list

Upvotes: 0

Views: 123

Answers (4)

Ajax1234
Ajax1234

Reputation: 71451

You can use list comprehension and enumerate at each iteration:

l = [[0,0,0,'P'],[0,0,0,0],[0,'Q',0,0],[0,0,0,0]]

indexes = [[i for i, a in enumerate(b) if a == "P" or a == "Q"] for b in l]

You can also use a dictionary that stores the letter and a corresponding list with the sublist index and the letter index in the sublist:

new_dict = [{a:[i, b] for b, a in enumerate(c) if a == "P" or a == "Q"} for i, c in enumerate(l)]

new_dict = [i for i in new_dict if i]

final_dict = {**new_dict[0], **new_dict[1]}
print(final_dict)

Output:

{'Q': [2, 1], 'P': [0, 3]}

Finally, to get the complete sublists:

final_values = final_dict.values()

Output:

[[2, 1], [0, 3]]

Upvotes: 1

MarianD
MarianD

Reputation: 14141

You have to search in nested lists:

for nested_list in list:
    if 'P' in nested_list:
        index_of_P = nested_list.index('P')
        break

The explanation:

index() method raises the ValueError when 'P' is not found in nested_list, hence the if test just before its use.

As you have only one 'P', break the for loop after the successful search.

Note:

This is a simple but not recommended usage. After learning about the exceptions handling you will see more appropriate solution.

Upvotes: 0

morfioce
morfioce

Reputation: 1117

Here the elements of your search array are themselves arrays so you need to do some work to get what you want.

I will suppose that when searching for 'P' and 'Q' you are expecting to get a tuple as a result where the first value refers to the array where the search element appears and the second value refers to the index of the search element in the array where it appears.

You can write the following function

def find_position(array, elem):
    for i, arr in enumerate(array):
        if elem in arr:
            return (i, arr.index(elem))
    return (-1, -1) # elem do not exist


find_position(l, 'Q')
Out: (2, 1)

find_position(l, 'P')
Out (0, 3)

Upvotes: 0

Yohan Grember
Yohan Grember

Reputation: 607

If you are seeking the index at which the element is in the sublist, you can try this:

input_list = [[0,0,0,'P'],[0,0,0,0],[0,'Q',0,0],[0,0,0,0]]

def find_position(input_list, element):
    index = 0
    for sublist in input_list:
        if element in sublist:
            return index
        index += 1
    return None

find_position(input_list, 'P')
OUT: 0

find_position(input_list, 'Q')
OUT: 2

Upvotes: 0

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