Making strings in a list equal to ints in a different list?

Question

If I have for instance, the list: ['a', 'b', 'c', 'd', 'e']

and I want each letter by index to be equal to the ints by index in this list:

[1, 1, 2, 1, 1]

so that 'a' == 1, 'b' == 1, 'c' == 2, etc.

Is there a simple way of doing this that I'm missing?

You mean you want to define those letters as variables? – zondo Mar 09 '16 at 00:50 — zondo, Mar 09 '16 at 00:50

score 1 · Answer 1 · answered Mar 09 '16 at 00:55

1

You can use a dictionary to hold the numeric values as values of the keys 'a' - 'e'.

You can set this up manually like this:

d = {'a': 1, 'b': 1, 'c': 2, 'd': 1, 'e': 1}

...or if you have the two lists already made...

d = dict()
for i, letter in enumerate(letterList):
    d[letter] = numberList[i]

answered Mar 09 '16 at 00:55

m_callens

1

You can make it one line with `d = {letter:numberList[i] for i, letter in enumerate(letterList)}` – zondo Mar 09 '16 at 00:56
2

or use `d = dict(zip(letter_list, number_list))` – mdxs Mar 09 '16 at 00:59

score 0 · Answer 2 · answered Mar 09 '16 at 01:56

The simplest way to do this would be to use a dictionary instead of a list. As a few others have suggested.

But if you absolutely need to use a list, I imagine you would want to just compare the values of the index after you find them.

alpha = ['a','b','c','d','e']
numeric = [1,1,2,1,1]
alpha.index('a') == numeric.index('1')
alpha.index('b') == numeric.index('1',1)

Note: The index method finds the first occurrence of the parameter so you need to pass it a starting point for duplicate items.

2 Answers2