Python replacing values using index and/or values

Question

consider:

a = [1, 1, 0, 1, 2]
b = [3, 2, 2, 0, 0]

I have to make take all items in list b that correspond to the value of 1 in the list a. Then change all occurrences to those items in both lists to 0.

Important Edit: I also have to take these items as input.

I have to also do this simultaneously for all items.

I have tried using a for loop to do this but for loop goes through the list one by one instead of all together.

I am honestly confused on how to approach this. Any help is appreciated! Thanks! NOTE: Has to be python 3.X

Super Edit: (Sorry I didn't do this earlier)

I want the code to go something like this:

input:

a = [1, 1, 0, 1, 2]
b = [3, 2, 2, 0, 0]
count = 0

output:

a = [0, 0, 0, 0, 0]
b = [0, 0, 0, 0, 0]
count = 1

where count is supposed to increment as the numbers are changed. Thus I have to go through the lists and do the steps described above simultaneously for all items and lists. Sorry I didn't put this up earlier.

• hygull: Can you please write the expected output?
• tafaust: The expected output should read a = [1, 1, 0, 1, 0] and b = [0, 0, 0, 0, 0] if I am not mistaken
• tafaust: @Grimlock How do you imagine the procedure to achieve your goal should lookalike?
• Grimlock: @Thomas I have updated the explanation
• Grimlock: @hygull I have written out the output now

Answer 1

if you want to do what you are asking and are looking for speed this will work:

[0 if x in b else x for x in a]

This will work, I am sure but slower than the above approach. But this will help in better understanding.

newa=[]
for i in a:
 if i in b:
 newa.append(0)
 else:
 newa.append(i)

the newa will be :

[1, 1, 0, 1, 0]

• Grimlock: The only problem with that is that the for loop goes through the items one by one. I need a way to go through them all simultaneously
• tafaust: @Grimlock Why has it to happen simultaneous?
• Grimlock: @inder can you write that without list comprehension
• Inder: @Grimlock I am sorry I don't think I follow, without list comprehension meaning ???
• tafaust: @Grimlock I don't think that what you request for is actually feasible... like list-comprehensions in python is "the thing" to do... somehow you need to traverse the list as there is no such thing as a "one-shot-solution" to this.

Answer 2

A first solution, using a loop:

def with_loop(a, b):
 out = []
 for index, val in enumerate(a):
 if val == 1:
 out.append(b[index])
 a[index] = 0
 b[index] = 0
 return out
a = [1, 1, 0, 1, 2]
b = [3, 2, 2, 0, 0]
print(with_loop(a, b))
# [3 2 0]
print(a)
# [0 0 0 0 2]
print(b)
# [0 0 2 0 0]

Some timings:

a = [1, 1, 0, 1, 2]
b = [3, 2, 2, 0, 0]
%timeit with_loop(a, b)
# 1000000 loops, best of 3: 806 ns per loop
a = [1, 1, 0, 1, 2]*20
b = [3, 2, 2, 0, 0]*20
%timeit with_loop(a, b)
#100000 loops, best of 3: 7.3 μs per loop
a = [1, 1, 0, 1, 2]*400
b = [3, 2, 2, 0, 0]*400
%timeit with_loop(a, b)
# 10000 loops, best of 3: 165 μs per loop

If your lists are large, it could be good to use numpy:

import numpy as np
a = np.array([1, 1, 0, 1, 2])
b = np.array([3, 2, 2, 0, 0])
def with_numpy(a, b):
 out = b[a == 1]
 b[a == 1] = 0
 a[a == 1] = 0
 return out
print(with_numpy(a, b))
# [3 2 0]
print(a)
# [0 0 0 0 2]
print(b)
# [0 0 2 0 0]

Again, some timings:

a = np.array([1, 1, 0, 1, 2])
b = np.array([3, 2, 2, 0, 0])
%timeit with_numpy(a, b)
#100000 loops, best of 3: 5.02 μs per loop
a = np.array([1, 1, 0, 1, 2]*20)
b = np.array([3, 2, 2, 0, 0]*20)
%timeit with_numpy(a, b)
#100000 loops, best of 3: 5.34 μs per loop
a = np.array([1, 1, 0, 1, 2]*400)
b = np.array([3, 2, 2, 0, 0]*400)
%timeit with_numpy(a, b)
#100000 loops, best of 3: 13.9 μs per loop

For small lists, using numpy has a big overhead, but it gets faster than the pure Python loops solution with lists less than 100 items long, and is more than 10 times faster for 2000 items.

• cs95: This is pretty comprehensive but you might've just succeeded in overwhelming some kid trying to get his homework done. Not that there's a problem with that, but eh...
• Grimlock: alright so my lists are never going to be too long so I'll go with the first solution. However, if there is a 1 in list a and 0 in list b I have to go back and turn the item in list b that corresponds to 0 in list a.

Answer 3

You could use a generator which inspects both sequences and yields their current and future values:

def gen(a,b):
 for i in len(a):
 if a[i]==1:
 yield b[i] # need some cleverness about how we unpack this
 yield (0,0)
 else:
 yield None
 yield (a[i],b[i])

Then something like assign a,b = list(gen(a,b)). We'll need extra code to unpack the current nonzero elements of b

Answer 4

Considering that you want to remove the 1s from list a too, please have a look at this solution proposal:

>>> x, y = map(list, zip(*[(0 if i==1 else i, 0 if i==1 else j) for i, j in zip(a, b)]))
>>> print('x: {}, y: {}'.format(x, y))
x: [0, 0, 0, 0, 2], y: [0, 0, 2, 0, 0]

Please keep in mind that there actually is a need to traverse the list and that there is no other way to do this except going through the list one by one.