I would like to know if there are any other ways to loop and manipulate data located in separate arrays.
import numpy as np
a = np.arange(2)
b = np.arange(5)
c = np.arange(5)
l1 = []
for x in a:
l2 = []
for y in b:
l3 = []
y = x + 1
for z in c:
z = x + y
t = (x,y,z)
l3.append(t)
l2.append(l3)
l1.append(l2)
print l1
This code does exactly what you are doing.
def method(lst, range1, range2):
for i in lst:
yield [[(i, i+1, 1+(i*2))]*range2]*range1
Can even be turned into a generator expression:
def gen_method(lst, r1, r2):
return ([[(i, i+1, 1+(i*2))]*r2]*r1 for i in lst)
Test it yourself if you like.
My tests:
a = range(2)
b = range(5)
c = range(5)
def your_method(a, b, c):
l1 = []
for x in a:
l2 = []
for y in b:
l3 = []
y = x + 1
for z in c:
z = x + y
t = (x,y,z)
l3.append(t)
l2.append(l3)
l1.append(l2)
return l1
def my_method(lst, range1, range2):
for i in lst:
yield [[(i, i+1, 1+(i*2))]*range2]*range1
yours = your_method(a, b, c)
mine = list(my_method(a, len(b), len(c)))
print yours
print '==='
print mine
print '==='
print yours == mine
>>>
[[[(0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1)], [(0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1)], [(0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1)], [(0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1)], [(0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1)]], [[(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)], [(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)], [(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)], [(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)], [(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)]]]
===
[[[(0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1)], [(0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1)], [(0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1)], [(0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1)], [(0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1)]], [[(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)], [(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)], [(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)], [(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)], [(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)]]]
===
True
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