Question

How do you compare a matrix row (ie list of lists) with a given string?

index = 99999
for i in range(len(text)):
    if (matrix[i][0:len(text)] == text):
        index = i

I want “index” to be the number of row for which “row == text”, but the above code outputs 99999.

I know for sure that exactly one of the rows contains the string.
For example, the matrix is

 ['a', 'i', 'n', 'e', 'm']
 ['e', 'm', 'a', 'i', 'n']
 ['i', 'n', 'e', 'm', 'a']
 ['m', 'a', 'i', 'n', 'e']
 ['n', 'e', 'm', 'a', 'i']

and I want to know which row is “maine” (number 3 in this case).
Thanks!

Answer 1

Try

try:
    index = matrix.index(list(text))
except IndexError:
    index = 99999

list(text) turns the string into a list of characters. list.index searches for the item you specify (using == as the equality comparison) and returns its index if found, or raises IndexError if not found.

I’d also not recommend using 99999 as the ‘not found’ value; it’s rather safer to use a value like -1 or (better yet) just leave the exception alone unless you intend to handle it.

If you know the string must be in the matrix, then index = matrix.index(list(text)) is all you need.