If I have a x list that contains multiple values, and a few repeats, and I want to have the index for each element, how can I do it? I tried doing it using the .index() method but it only returns the index of the first element.
>>> x = ['c','a','s','a']
>>> a.index('a')
1
But what I am trying to do is to have the index of each letter, even having two numbers for the repeated letters and move to a dictionary:
x_index = {"c":0,"a":(1,3),"s":2}
You can use enumerate to iterate over the list and get the index of repeated elements:
>>> lista = ['c', 'a', 's', 'a']
>>> [i for i, item in enumerate(lista) if item == 'a']
[1, 3]
>>>
To save the repeated values and indexes in a dictionary , first check that the key already exists, if so, you get the repeated indexes and update the key, otherwise just enter the value:
dicionario = {}
lista = ['c', 'a', 's', 'a', 's']
for valor in lista:
if valor in dicionario:
dicionario[valor] = [i for i, item in enumerate(lista) if item == valor]
else:
dicionario[valor] = lista.index(valor)
print (dicionario)
# {'a': [1, 3], 'c': 0, 's': [2, 4]}
The dictionary order will not always be the same, if you need the order preserved, use collections.OrderedDict :
from collections import OrderedDict
dicionario = OrderedDict()
# ...
print (dicionario)
# OrderedDict([('c', 0), ('a', [1, 3]), ('s', [2, 4])])