Python loop counter in a for loop [duplicate]

Answer 1

enumerate is what you are looking for.

You might also be interested in unpacking:

# The pattern
x, y, z = [1, 2, 3]

# also works in loops:
l = [(28, 'M'), (4, 'a'), (1990, 'r')]
for x, y in l:
    print(x)  # prints the numbers 28, 4, 1990

# and also
for index, (x, y) in enumerate(l):
    print(x)  # prints the numbers 28, 4, 1990

Also, there is itertools.count() so you could do something like

import itertools

for index, el in zip(itertools.count(), [28, 4, 1990]):
    print(el)  # prints the numbers 28, 4, 1990

Answer 2

You could also do:

 for option in options:
      if option == options[selected_index]:
           #print
      else:
           #print

Although you'd run into issues if there are duplicate options.

Answer 3

Use enumerate() like so:

def draw_menu(options, selected_index):
    for counter, option in enumerate(options):
        if counter == selected_index:
            print " [*] %s" % option
        else:
            print " [ ] %s" % option    

options = ['Option 0', 'Option 1', 'Option 2', 'Option 3']
draw_menu(options, 2)

Note: You can optionally put parenthesis around counter, option, like (counter, option), if you want, but they're extraneous and not normally included.

Answer 4

I'll sometimes do this:

def draw_menu(options, selected_index):
    for i in range(len(options)):
        if i == selected_index:
            print " [*] %s" % options[i]
        else:
            print " [ ] %s" % options[i]

Though I tend to avoid this if it means I'll be saying options[i] more than a couple of times.