In the code below, I can easily reduce the array ['a','b','a','c','b','b','c','a'] to a binary array [0 1 0 1 1 1 1 0] so that 'a' -> 0 and 'b','c' -> 1. How do I transform it to a ternary array so that 'a' -> 0, 'b' -> 1, 'c' -> 2, without using for and if-else? Thanks.
import numpy as np
x = np.array(['a', 'b', 'a', 'c', 'b', 'b', 'c', 'a'])
y = np.where(x=='a', 0, 1)
print(y)
By doing:
np.where(x == 'a', 0, (np.where(x == 'b', 1, 2)))
note that this changes all the characters that are neither 'a' or 'b' to 2. I've assumed that you have only an array with a,b and c.
A more scalable version is using dictionary of conversion:
my_dict = {'a':0, 'b':1, 'c':2}
x = np.vectorize(my_dict.get)(x)
output:
[0 1 0 2 1 1 2 0]
Another approach is:
np.select([x==i for i in ['a','b','c']], np.arange(3))
For small dictionary @ypno's answer is going to be faster. For larger dictionary, use this answer.
Time Comparison:
Ternary alphabet:
lst = ['a','b','c']
my_dict = {k: v for v, k in enumerate(lst)}
#@Ehsan's solution1
def m1(x):
return np.vectorize(my_dict.get)(x)
#@ypno's solution
def m2(x):
return np.where(x == 'a', 0, (np.where(x == 'b', 1, 2)))
#@SteBog's solution
def m3(x):
y = np.where(x=='a', 0, x)
y = np.where(x=='b', 1, y)
y = np.where(x=='c', 2, y)
return y.astype(np.integer)
#@Ehsan's solution 2 (also suggested by user3483203 in comments)
def m4(x):
return np.select([x==i for i in lst], np.arange(len(lst)))
#@juanpa.arrivillaga's solution suggested in comments
def m5(x):
return np.array([my_dict[i] for i in x.tolist()])
in_ = [np.random.choice(lst, size = n) for n in [10,100,1000,10000,100000]]
Same analysis for 8 letter alphabet:
lst = ['a','b','c','d','e','f','g','h']
