Python : printing number sequences from nested loops

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I know it seems very basic and easy, and I'm sorry if I'm wasting everybody's time, but I'm stuck and it's been hours now.

I have 4 loops, and in the end I must print all the numbers from 1 to the end. It will be moe clear after you see the code:

for a in range (0,7):
    for b in range (1,118):
        for c in range (0,7):
            for d in range (1,118):
                print(...)

What I expect is :

1
2
3
...
383292

where 383292=118*118*7*4

Keep in mind that I really need these 4 loops as they are shown.

What should I put inside the print? I feel so stupid right now because it seems to be a simple math problem but I can't seem to figure out the solution. Thanks a lot.

EDIT: to answers to all the comments, let's say I have this:

for a in range (0,2):
    for b in range (1,3):
        for c in range (0,2):
            for d in range (1,3):
                print(...)

I want to have as output:

1
2
3
...
24

I need to find some operations to do on a,b,c and d to have my results. I hope it is a bit clearer.

(Yeah, the counter may be easier in most cases, but this is cooler!)

You need to know the amount of outputs that occur per iteration of each loop.

Assuming your loops have this many cycles:

cycles_a = 7
cycles_b = 117
cycles_c = 4
cycles_d = 117

You have outputs:

outputs_d = 1
outputs_c = outputs_d * cycles_d
outputs_b = outputs_c * cycles_c
outputs_a = outputs_b * cycles_b

If you run this loop:

for a in range(0,cycles_a):
    for b in range(0,cycles_b):
        for c in range(0,cycles_c):
            for d in range(0,cycles_d):
                print(d*outputs_d + c*outputs_c + b*outputs_b + a*outputs_a)

You will print:

0
1
...
<7*117*4*117 - 1>

If you want to start by printing 1, add one to the formula.

If your ranges start at 1 instead of 0, use (a-1)/(b-1)/(c-1)/(d-1) in the formula.

python nested loop hailstone sequence for range, Your program will print out the number that has the largest cycle length and what that cycle length is. Your sample session will look like this: Enter starting number � The for loop prints the number from 1 to 10 using the range () function here i is a temporary variable which is iterating over numbers from 1 to 10. It’s worth mentioning that similar to list indexing in range starts from 0 which means range ( j ) will print sequence till ( j-1) hence the output doesn’t include 6.

First, there is the straightforward and direct way:

counter = 1
for a in range (0,7):
    for b in range (1,118):
        for c in range (0,7):
            for d in range (1,118):
                print(counter)
                counter += 1

If we want to not do that, then we might note that it's as though you are writing a number in a mixed base (as opposed to base 10). It sort of seems like your number is in the form

A B C D
| | | |__ base 118
| | |____ base 7
| |______ base 118
|________ base 7

To convert a number from this mixed base to base 10, you do

def convert(a, b, c, d):
    return d + 7 * c + 118 * 7 * b + 7 * 118 * 7 * a

(Or you reverse the digits and endian-ness).

So then you might do

for a in range (0,7):
    for b in range (118):
        for c in range (0,7):
            for d in range (118):
                print(d + 7 * c + 118 * 7 * b + 7 * 118 * 7 * a)

1.13. Loops and Sequences — Hands-on Python Tutorial for Python 3, There are two Python statement types to do that: the simpler for loops, which we take up def tripleAll(nums): ''' print triple each of the numbers in the list nums. I am new to python and I am having some problems with this code that should print all the prime numbers that are smaller than 50 using nested loops. Here is the code: i = 2 while(i &lt; 50):

A simple solution (in the format you are doing it) would be:

counter = 0 
myrange = lambda x: range(1,x+1)

for a in myrange(118):
    for b in myrange(118):
        for c in myrange(7):
            for d in myrange(4):
                counter+= 1 
                print('counter %d gives: %d = %d * %d * %d * %d'%(counter,a*b*c*d,a,b,c,d))


myrange runs a command starting at 1 and ending at x < maxnum+1 (ie. your max number).

The print uses cstyle formatting where %d is an integer %s is a string etc and subs in the arguments after %.

This produces:

counter 1 gives: 1 = 1 * 1 * 1 * 1

...

counter 389869 gives: 97468 = 118 * 118 * 7 * 1
counter 389870 gives: 194936 = 118 * 118 * 7 * 2
counter 389871 gives: 292404 = 118 * 118 * 7 * 3
counter 389872 gives: 389872 = 118 * 118 * 7 * 4

Alternatively, you can save the values in a list and print them one at a time.

myrange = lambda x: range(1,x+1)
output=[]
for a in myrange(118):
    for b in myrange(118):
        for c in myrange(7):
            for d in myrange(4):
                output.append(a*b*c*d)

for i in output:
   print(i)

This gives:

1
...
250632
334176
97468
194936
292404
389872

If you need the values to be sored in size, you can do

output.sort()

for i in output:
   print(i)

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I guess that the easier way is:

res = 0
for a in range (0,2):
    for b in range (1,3):
        for c in range (0,2):
            for d in range (1,3):
                res += 1
                print(res)

but it prints a max of 16 numbers, not 24. You should start counting from 0 to n, or instead of (1, 3) you should write (1,4) if you want your cycle to repeat 3 times.

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Another way to do it is using a generator:

>>> def infinite_counter(n=0):
        while True:
            n += 1
            yield n

>>> counter = infinite_counter()

>>> for i in range(3):
       for j in range(2):
           print(next(counter))
1
2
3
4
5
6

The itertools.product is even better if you don't really need the ugly 4+ levels of nested for loops (if you are discarding the values in the iterables and just want the counting):

>>> for i, _ in enumerate(product(range(3), range(2))):
        print(i+1)
1
2
3
4
5
6

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Comments
  • Do you want to print the number just one time? or multiple times?
  • Maybe start with a small range, and provide input and output ?
  • Not quite clear from your question but do you just need to declare e = 1 at the start, print e, and then e += 1 after the print?
  • If I understood you correctly, you can put print statement below each for loop declaration, but then you'll have duplicated values. Another thing is the question, if you need these values to be ordered or not. Please provide clear input and output as @hansolo suggested.
  • Did you try a simple "counter" variable like David Buck suggested? Your math was wrong -- you will print 16 times because 16 == 2*2*2*2, which is the ranges you selected. You edited out the multiplication which shows your intent.
  • That's exacctly what I wanted, Thanks a lot
  • Now that's a clever way of getting around the ambiguities of the question. The counter won't end on the right number, but the second part of this answer will actually answer the original question, taking some liberties with the elipses to cover up problems in the original problem statement, and meet the "function of a,b,c,d" stipulation. You may have some math errors and start and end on the wrong numbers.