Python：如何在重复的列表中始终找到下一个索引？

Question

Sorry if the question wasn't clear enough, I am very new to python. I also apologize in advance if there are any typos in my code.

Say I have a list

list = [a,b,c,a,x,y,b,m,a,z]

And I want to get the index value of the element after each 'a' using a for loop and store it in a dict. (This assumes dict = {} already exists)

for store in list:
    if dict.has_key(store) == False:
        if list.index(store) != len(list)-1:
            dict[store] = []
            dict[store].append(list[list.index(store)+1])
    else:
        if list.index(store) != len(list)-1:
            dict[store].append(list[list.index(store)+1])

Now ideally, I would want my dict to be

dict = {'a':['b','x','z'], 'b':['c','m'], 'c':['a']....etc.}

Instead, I get

dict = {'a':['b','b','b'], 'b':['c','c'], 'c':['a']...etc.}

I realized this is because index only finds the first occurrence of variable store. How would I structure my code so that for every value of store I can find the next index of that specific value instead of only the first one?

Also, I want to know how to do this only using a for loop; no recursions or while, etc (if statements are fine obviously).

I apologize again if my question isn't clear or if my code is messy.

Answer 1

You can do it like that:

l = ['a','b','c','a','x','y','b','m','a','z']
d={}
for i in range(len(l)-1):
    if not l[i] in d:
        d[l[i]] = []
    d[l[i]].append(l[i+1])

Then d is

{'a': ['b', 'x', 'z'],
 'b': ['c', 'm'],
 'c': ['a'],
 'm': ['a'],
 'x': ['y'],
 'y': ['b']}

Regarding your code, there is no need to use index , as you already enumerating over the list, so you do not need to search for the place of the current element. Also, you can just enumerate until len(l)-1 , which simplifies the code. The problem in your code was that list.index(store) always finds the first appearance of store in list .

Answer 2

This looks like a job for defaultdict . Also, you should avoid using list and dict as variables since they are reserved words.

from collections import defaultdict

# create a dictionary that has default value of an empty list
# for any new key
d = defaultdict(list)

# create the list
my_list = 'a,b,c,a,x,y,b,m,a,z'.split(',')

# create tuples of each item with its following item
for k,v in zip(my_list, my_list[1:]):
    d[k].append(v)

d 
# returns:
defaultdict(list,
            {'a': ['b', 'x', 'z'],
             'b': ['c', 'm'],
             'c': ['a'],
             'm': ['a'],
             'x': ['y'],
             'y': ['b']})