I have two list like this:
>>> a = ['a', 'b', 'a', 'c', 'b', 'a', 'd'] >>> b = ['a', 'b', 'c', 'd'] By using b I want to get the result like this:
a -> 0, 2, 5 b -> 1, 4 c -> 3 d -> 6 Tried using enumerate()
>>> for i, j in enumerate(b): ... a[i] ... 'a' 'b' 'a' 'c' Didn't work.
You were right to use enumerate, though you didn't quite use it exactly right
In [5]: a = ['a', 'b', 'a', 'c', 'b', 'a', 'd'] In [6]: b = ['a', 'b', 'c', 'd'] In [7]: for char in b: print char, [i for i,c in enumerate(a) if c==char] a [0, 2, 5] b [1, 4] c [3] d [6] 
>>> [(char, [i for i,c in enumerate(a) if c==char]) for char in b] [('a', [0, 2, 5]), ('b', [1, 4]), ('c', [3]), ('d', [6])] or
>>> dict((char, [i for i,c in enumerate(a) if c==char]) for char in b) {'a': [0, 2, 5], 'c': [3], 'b': [1, 4], 'd': [6]} 
def get_all_indexes(lst, item): return [i for i, x in enumerate(lst) if x == item] a = ['a', 'b', 'a', 'c', 'b', 'a', 'd'] b = ['a', 'b', 'c', 'd'] for item in b: print item, get_all_indexes(a, item) Result:
>>> a [0, 2, 5] b [1, 4] c [3] d [6] I ll do it like...
Code:
a = ['a', 'b', 'a', 'c', 'b', 'a', 'd'] b = ['a', 'b', 'c', 'd'] for item in b: print item + ':' + ','.join([str(i) for i,val in enumerate(a) if item==val]) Output:
a:0,2,5 b:1,4 c:3 d:6 Hope this helps :)
Try
>>> a = ['a', 'b', 'a', 'c', 'b', 'a', 'd'] >>> b = ['a', 'b', 'c', 'd'] >>> indices = [ i for i, x in enumerate(a) if x == b[0] ] >>> indices [0, 2, 5] You can change b[0] to whatever letter you want the indices for.
Explanation:
Let each element of enumerate(a) be of the form (i,x). So, indices is an array of all i in enumerate(a) such that x equals b[0] (or whatever other letter you want it to be).
import collections def getIndex(ListA, ListB): res = collections.defaultdict(list) for element in ListB: for (i, v) in enumerate(ListA): if element == v: res[v].append(i) for key, value in sorted(res.items(), key = lambda d : d[0]): print(key, " -> ", ",".join([str(i) for i in value])) a = ['a', 'b', 'a', 'c', 'b', 'a', 'd'] b = ['a', 'b', 'c', 'd'] getIndex(a, b) The output is：
a -> 0,2,5 b -> 1,4 c -> 3 d -> 6 Thanks for all the enumerate :), here is something pretty interesting I found last time trying to answer something quite the same, that is to use index() function to return multiple positions.
a = ['a', 'b', 'a', 'c', 'b', 'a', 'd'] b = ['a', 'b', 'c', 'd'] for item in b: index = [] start = -1 while True: #also take care of the case where item in b is not in a try: start = a.index(item, start+1) index.append(start) except ValueError: break; print item, index Result
a [0, 2, 5] b [1, 4] c [3] d [6] a.index(b, position) defines the starting point of indexing.
O(n+m) algorithm (many other answers seem to have O(n*m) complexity):
>>> from collections import defaultdict >>> D = defaultdict(list) >>> for i,item in enumerate(a): D[item].append(i) >>> for item in b: print('{} -> {}'.format(item, D[item])) a -> [0, 2, 5] b -> [1, 4] c -> [3] d -> [6] 
[map(lambda pair: pair[0], filter(lambda x: x[1] == i, enumerate(a))) for i in b]