Python simplifying nested for loops using recursive function or other methods

Question

I would like to simplify the following codes:

import numpy as np
interval = 20
wgt = list(np.arange(0, 101, interval))

pairs = []
for a in wgt:
    for b in list(np.arange(0, 101-a, interval)):
        for c in list(np.arange(0, 101-a-b, interval)):
            for d in list(np.arange(0, 101-a-b-c, interval)):
                for e in list(np.arange(0, 101-a-b-c-d, interval)):
                    for f in list(np.arange(0, 101-a-b-c-d-e, interval)):
                        for g in list(np.arange(0, 101-a-b-c-d-e-f, interval)):
                            for h in list(np.arange(0, 101-a-b-c-d-e-f-g, interval)): 
                                for i in list(np.arange(0, 101-a-b-c-d-e-f-g-h, interval)):
                                    j = 100-a-b-c-d-e-f-g-h-i
                                    pairs.append([a,b,c,d,e,f,g,h,i,j])

Eventually, I want to obtain the pairs repeating for loops N times. The number of columns in pairs[] increases with the number of loops.

Can somebody possibly simplify the above code? I know one possible solution is using recursive function, but it is too challenging task for me, beginner. I do not care if your code includes other methods or syntax as long as it simplifies codes.

Thank you in advance!

Try to reduce the problem down to its most basic form.

ohailolcat
– ohailolcat

2020-11-01 14:09:18 +00:00
Commented Nov 1, 2020 at 14:09 — ohailolcat
– ohailolcat, Commented Nov 1, 2020 at 14:09

Alex P · Accepted Answer · 2020-11-01 13:35:39Z

2

The solution is the following recursive function:

def foo(n, wgt, s):
  if n==1:
    return [[100-s]]
  
  pairs = []
  for w in wgt:
    if s+w > 100: continue
    for t in f(n-1, wgt, s+w):
      pairs.append([w] + t)
  return pairs

And you can produce the desired pairs list with:

import numpy as np
interval = 20
wgt = np.arange(0, 101, interval)
N = 10

pairs = foo(N, wgt, 0)

answered Nov 1, 2020 at 13:35

Alex P

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