I have a list of strings. I want to assign a unique number to each string (the exact number is not important), and create a list of the same length using these numbers, in order. Below is my best attempt at it, but I am not happy for two reasons:
It assumes that the same values are next to each other
I had to start the list with a 0, otherwise the output would be incorrect
My code:
names = ['ll', 'll', 'll', 'hl', 'hl', 'hl', 'LL', 'LL', 'LL', 'HL', 'HL', 'HL']
numbers = [0]
num = 0
for item in range(len(names)):
if item == len(names) - 1:
break
elif names[item] == names[item+1]:
numbers.append(num)
else:
num = num + 1
numbers.append(num)
print(numbers)
I want to make the code more generic, so it will work with an unknown list. Any ideas?
Without using an external library (check the EDIT for a Pandas solution) you can do it as follows :
d = {ni: indi for indi, ni in enumerate(set(names))}
numbers = [d[ni] for ni in names]
Brief explanation:
In the first line, you assign a number to each unique element in your list (stored in the dictionary d; you can easily create it using a dictionary comprehension; set returns the unique elements of names).
Then, in the second line, you do a list comprehension and store the actual numbers in the list numbers.
One example to illustrate that it also works fine for unsorted lists:
# 'll' appears all over the place
names = ['ll', 'll', 'hl', 'hl', 'hl', 'LL', 'LL', 'll', 'LL', 'HL', 'HL', 'HL', 'll']
That is the output for numbers:
[1, 1, 3, 3, 3, 2, 2, 1, 2, 0, 0, 0, 1]
As you can see, the number 1 associated with ll appears at the correct places.
EDIT
If you have Pandas available, you can also use pandas.factorize (which seems to be quite efficient for huge lists and also works fine for lists of tuples as explained here):
import pandas as pd
pd.factorize(names)
will then return
(array([(array([0, 0, 1, 1, 1, 2, 2, 0, 2, 3, 3, 3, 0]),
array(['ll', 'hl', 'LL', 'HL'], dtype=object))
Therefore,
numbers = pd.factorize(names)[0]
If the condition is that the numbers are unique and the exact number is not important, then you can build a mapping relating each item in the list to a unique number on the fly, assigning values from a count object:
from itertools import count
names = ['ll', 'll', 'hl', 'hl', 'LL', 'LL', 'LL', 'HL', 'll']
d = {}
c = count()
numbers = [d.setdefault(i, next(c)) for i in names]
print(numbers)
# [0, 0, 2, 2, 4, 4, 4, 7, 0]
You could do away with the extra names by using map on the list and a count object, and setting the map function as {}.setdefault (see @StefanPochmann's comment):
from itertools import count
names = ['ll', 'll', 'hl', 'hl', 'LL', 'LL', 'LL', 'HL', 'll']
numbers = map({}.setdefault, names, count()) # call list() on map for Py3
print(numbers)
# [0, 0, 2, 2, 4, 4, 4, 7, 0]
As an extra, you could also use np.unique, in case you already have numpy installed:
import numpy as np
_, numbers = np.unique(names, return_inverse=True)
print(numbers)
# [3 3 2 2 1 1 1 0 3]
If you have k different values, this maps them to integers 0 to k-1 in order of first appearance:
>>> names = ['b', 'c', 'd', 'c', 'b', 'a', 'b']
>>> tmp = {}
>>> [tmp.setdefault(name, len(tmp)) for name in names]
[0, 1, 2, 1, 0, 3, 0]