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?
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)
mylist is long.
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:
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
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.
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]
You can implement your own findIndex function, this looks like Javascipt's Array.prototype.findIndex
def findIndex(callback, list):
for index, element in enumerate(list):
if callback(element):
return index
return -1
A little bit of "amateur" approach and yet very easy to understand:
Creating a flag boolean value that will execute the result based on if it is True or False with a nested for loop.
entry=input()
listtt=[["a","b","c"],["d","e","f"],["g","h"]]
found=False
for sublist in listtt:
for charr in sublist:
if entry==charr:
print(f"'{entry}' found in sublist {listtt.index(sublist)} with index {sublist.index(charr)}")
found=True
break
if found==False:
print(f"'{entry}' not found")
Expanding on DainDwarf's answer, this will return multiple values if the searched item is in more than one of the nested lists:
def find_in_list_of_list(mylist, char):
found_list = []
for sub_list in mylist:
if char in sub_list:
found_list.append(f'({mylist.index(sub_list)}, {sub_list.index(char)})')
if len(found_list) > 0:
return found_list
example_list = [['a', 'b', 'c'], ['d', 'e', 'f'], ['g', 'h'], ['a', 'b', 'c', 'd']]
print(find_in_list_of_list(example_list, 'c'))
I also got en error when changing the search from 'b' to 'x' in Dwarf's code:
Traceback (most recent call last):
File "C:\main.py", line 9, in <module>
print(find_in_list_of_list(example_list, 'x'))
File "C:\main.py", line 5, in find_in_list_of_list
raise ValueError("'{char}' is not in list".format(char = char))
ValueError: 'x' is not in list
a