How to find the index of a list that contains an int in python

Question

If I have this list of lists

myList = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

and I would like to set myVar equal to the index of the list that contains the number 5. In the end myVar would equal 1. How would I do that? I am new to python so a simple answer would be appreciated.

Answer 1

myList = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
# loop if 5 in the sub_list, than give the index
myVar = [myList.index(ls) for ls in myList if 5 in ls]
print myVar 
# 1

Answer 2

Use for -loop to get sublist from myList and then check if 5 in sublist:

This way you find first row with 5

my_list = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
my_var = None

for number, row in enumerate(my_list):
    if 5 in row:
        my_var = number
        break # don't search in other rows

print(my_var)

if you need all rows with 5 then you list for results

my_list = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
my_var = [] # list for all results

for number, row in enumerate(my_list):
    if 5 in row:
        my_var.append( number )

print(my_var)

or in one line

my_list = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

my_var = [number for number, row in enumerate(my_list) if 5 in row]

print(my_var)

Answer 3

You may create a custom function using enumerate() to get the index as:

def get_index(my_list, val):
    for i, item in enumerate(my_list):
        if val in item:
            return i
    else:
        raise ValueError('{} not found in {}'.format(val, my_list))

Sample run:

>>> myList = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

# Valid value
>>> get_index(myList, 5)
1

# Invalid value
>>> get_index(myList, 22)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 6, in get_index
ValueError: 22 not found in [[1, 2, 3], [4, 5, 6], [7, 8, 9]]