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
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.
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)
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
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:
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: