Map unique strings to integers in Python [duplicate]

Answer 1

You'd use a hashmap in Python, too, but we call it a dict.

>>> L = ['apple','bat','apple','car','pet','bat']
>>> idx = 1
>>> seen_first = {}
>>>
>>> for word in L:
...     if word not in seen_first:
...         seen_first[word] = idx
...         idx += 1
... 
>>> [seen_first[word] for word in L]
[1, 2, 1, 3, 4, 2]

Answer 2

You can try:

>>> L = ['apple','bat','apple','car','pet','bat']
>>> l_dict = dict(zip(set(L), range(len(L))))
>>> print l_dict
{'pet': 0, 'car': 1, 'bat': 2, 'apple': 3}
>>> [l_dict[x] for x in L]
[3, 2, 3, 1, 0, 2]

Answer 3

You can use a map dictionary:

d = {'apple':1, 'bat':2, 'car':3, 'pet':4}
L = ['apple','bat','apple','car','pet','bat']
[d[x] for x in L] # [1, 2, 1, 3, 4, 2]

For auto creating map dictionary you can use defaultdict(int) with a counter.

from collections import defaultdict
d = defaultdict(int)
co = 1
for x in L:
    if not d[x]:
        d[x] = co
        co+=1
d # defaultdict(<class 'int'>, {'pet': 4, 'bat': 2, 'apple': 1, 'car': 3})

Or as @Stuart mentioned you can use d = dict(zip(set(L), range(len(L)))) for creating dictionary

Answer 4

x = list(set(L))
dic = dict(zip(x, list(range(1,len(x)+1))))

>>> [dic[v] for v in L]
[1, 2, 1, 3, 4, 2]

Answer 5

Lnew = []
for s in L:
    Lnew.append(hash(s))  # hash(x) returns a unique int based on string

Answer 6

Here's a quick solution:

l = ['apple','bat','apple','car','pet','bat']

Create a dict that maps all unique strings to integers:

d = dict([(y,x+1) for x,y in enumerate(sorted(set(l)))])

Map each string in the original list to its respective integer:

print [d[x] for x in l]
# [1, 2, 1, 3, 4, 2]