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.
np.select for more than a binary outcome.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']
np.vectorize? it really should be avoided. Almost certainly, np.array([my_dict[x] for x in array.tolist()]) will be faster.
np.vectorize seems to be faster for larger arrays than a list comprehension. I will add time comparison for better judgement. Thank you..tolist(), and then try it with with a regular list, and I bet you will see, it is faster with the list comprehension.