Finding element index in nested list

Question

Looking to find the element index of an element in a nested loop.

table = [
    [1, "Y", 3],
    [4, 5, "K"]
]      
for i in range(len(table[0])):
    for j in range(len(table)):
        print(table[i].index("Y"))`

But every time I run the the code it tells me Y is not in my list

`for t in table: if "Y" in t: print(t.index("Y"))` – RoadRunner Apr 17 '20 at 00:23 — RoadRunner, Apr 17 '20 at 00:23

score 3 · Answer 1 · answered Apr 17 '20 at 00:25

3

You don't need the nested loop. You can loop through the rows and call index() on the row. You can catch the error that will happen when Y is not in the list. This is a Python style of asking forgiveness instead of permission:

for row in table:
    try:
        print(row.index("Y"))
    except ValueError:
        pass

If you just want to know if 'Y' is in the table you can use any():

any('Y' in row for row in table)
# True

answered Apr 17 '20 at 00:25

Mark

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Sorry about that I didn't specify that I was looking for the column index! – SadTacoMadTaco Apr 17 '20 at 00:56
1

@SadTacoMadTaco `row.index("Y")` will give you the column index. – RoadRunner Apr 17 '20 at 00:57

Prune · Answer 2 · 2020-04-17T03:27:52.340

1

Your inner loop doesn't make any sense; index already covers the entire list; why do you need to do that 3 times?

You will get an error if the item doesn't appear. Instead

for i in range(len(table)):
    if 'Y' in table[i]:
        print(i, table[i].index('Y'))

For each row, you check whether Y appears. If it does, then you print the row and column number where it appears.

edited Apr 17 '20 at 03:27

answered Apr 17 '20 at 00:22

Prune

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1

`index["Y"]` is invalid. – RoadRunner Apr 17 '20 at 00:25
1

@Prune, I suggest you also add the first iteration in your answer for clarity – Dionys Apr 17 '20 at 00:25

Finding element index in nested list

2 Answers2