Create a loop for a set of similar statements in python

Question

I have a 2D matrix called A0

A0 = [[0 for x in range(3)] for y in range(3)]

I have a function called newProb which takes this as an argument and returns another 2D matrix in the following way:-

A1 = newProb(A0)

So, I want to put this code into a loop.

A1 = newProb(A0)
A2 = newProb(A1)
A3 = newProb(A2)
A4 = newProb(A3)

Any help will be appreciated. Thanks!

P.S. I have to make 100 calls to that function in the above way

Possible duplicate of [generating variable names on fly in python](http://stackoverflow.com/questions/4010840/generating-variable-names-on-fly-in-python) — Jean-François Fabre, Dec 05 '16 at 22:11
Why not just create a list, add the result of each call to the list, then on the next iteration use the previous list value? Or you could use recursion. — SteveJ, Dec 05 '16 at 22:14

score 1 · Answer 1 · edited Dec 05 '16 at 22:50

1

Rather than trying to create local variables in your loop, you'd be better off storing the results in a list. Your list's index will line up nicely with your naming convention.

A = []

A.append([[0 for x in range(3)] for y in range(3)])

A[1] = newProb(A[0])
A[2] = newProb(A[1])
# ...

You can then easily put this into a loop with a range.

# start the range off by 1
for i in range(1, 10):
    A[i] = newProb(A[i - 1])

This could also be written as

for i in range(10):
    A.append(newProb(A[i]))

edited Dec 05 '16 at 22:50

Nihal Rp

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answered Dec 05 '16 at 22:14

Soviut

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`A[1]` will give you `IndexError` if elements of `A` are not pre-allocated. Use `.append()` instead – jez Dec 05 '16 at 22:16
You mean A[0] instead of A0 in the second line ? – Nihal Rp Dec 05 '16 at 22:16
I realized. I've fixed the logic by offsetting the range by 1. – Soviut Dec 05 '16 at 22:17

score 0 · Answer 2 · answered Dec 05 '16 at 22:14

0

Use a dict:

A = {}
A[0] = [[0 for x in range(3)] for y in range(3)]
for i in range(1, 101):
    A[i] = newProb(A[i-1])

answered Dec 05 '16 at 22:14

jez

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score 0 · Answer 3 · answered Dec 05 '16 at 22:16

0

You can store your values in a list generated like so:

A = [A0]
for i in range(99):
    A.append(newProb(A[i]))

answered Dec 05 '16 at 22:16

Keiwan

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score 0 · Answer 4 · answered Dec 05 '16 at 22:16

0

Combining the answers from @jez and @Soviut:

A = []
A.append([[0 for x in range(3)] for y in range(3)])

for i in range(100):
    A.append(newProb(A[-1])

A[-1] is the last element in the list.

answered Dec 05 '16 at 22:16

Brad Campbell

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score 0 · Answer 5 · answered Dec 05 '16 at 22:18

0

If you just need the 'final' value, after the 100st loop:

value = [[0 for x in range(3)] for y in range(3)]
for i in range(100):
    value = newProb(value)

answered Dec 05 '16 at 22:18

Maurice Meyer

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score 0 · Answer 6 · answered Dec 05 '16 at 22:26

0

Round out the answers with a while loop

A = [A0]
while len(A) < 100:
    A.append(newProb(A[-1]))

answered Dec 05 '16 at 22:26

wwii

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Create a loop for a set of similar statements in python

6 Answers6