CODEWITHSUNDEEP
pythonpython-3.x
user10441782
user10441782

Reputation:

How do you use nested for loops to print out the following pattern in python?

How can you use nested for loops to print out the following pattern in python? So you don't have to write 7 print functions to print the pattern.

1

0 1

1 0 1

0 1 0 1

1 0 1 0 1

0 1 0 1 0 1

1 0 1 0 1 0 1

Upvotes: 3

Views: 1071

Answers (9)

piman314
piman314

Reputation: 5355

One more solution not using nested loops, instead using a single loop and binary representations.

solution = 1
for i in range(7):
    out_str = f'{solution:b}'
    print(' '.join(out_str[::-1]))
    solution = solution << 1
    if i > 0 and i % 2 == 1:
        solution |= 1

Upvotes: 0

Somesh
Somesh

Reputation: 72

You can try this :

    import os
    import re
    import sys

    def print_pattern(num):
        if(num == 1):
            print("1",end="\n")
            return
        elif(num%2 == 1):
            print("1",end=" ")
            print_pattern(num-1)
        else:
            print("0",end=" ")
            print_pattern(num-1)

    number = int(input("Enter the steps"))
    for i in range(1,number+1):
        print_pattern(i)

Upvotes: 0

Colonel Beauvel
Colonel Beauvel

Reputation: 31161

Not containing loop but an example of recursive function so that you push further:

>>> def foo(length_max, list_=[1]):
...     print(list_)
...     if len(list_)==length_max:
...         return
...     return foo(length_max, [int(not bool(list_[0]))] + list_)
...
>>> foo(7)
[1]
[0, 1]
[1, 0, 1]
[0, 1, 0, 1]
[1, 0, 1, 0, 1]
[0, 1, 0, 1, 0, 1]
[1, 0, 1, 0, 1, 0, 1]

Upvotes: 2

supertux
supertux

Reputation: 2179

This is the long but simple solution

count = -1 # Count where in the loop we are
switch = 1 # Switch between 0 and 1
start = 1  # Switch which varibale we start the line with
for i in range(8):
    print()
    count += 1
    start = switch
    for j in range(8):
        if j != count:
            if switch == 0:
                print("1 ", end='') #end= means dont print a new line character
                switch = 1
            else:
                print("0 ", end='')
                switch = 0
        if j == count:
            if start == 1:
                switch = 0
            else:
                switch = 1
            break #Break loop as we have printed as many characters as we want
print() # Print to end the line

Upvotes: 0

user3483203
user3483203

Reputation: 51165

This is not an answer in pure python, it makes use of numpy and scipy, but I think this is a great question because of the matrix you're trying to describe.

You may not be intending this, but your output matches the lower triangle of a Toeplitz matrix.

Because of the way you alternate, every diagonal will have the same value. Therefore, you could use scipy to produce your output.

Setup

num = 7
out = [1,0]*((num // 2)+1)
out = out[:num]

from scipy.linalg import toeplitz

res = toeplitz(out)
res[np.triu_indices_from(res, k=1)] = -1
for row in res:
    print(' '.join(map(str, row[row!=-1])))


1
0 1
1 0 1
0 1 0 1
1 0 1 0 1
0 1 0 1 0 1
1 0 1 0 1 0 1

Upvotes: 1

SpghttCd
SpghttCd

Reputation: 10860

As I can see, this is the contest of how many people do not answer the question, which was about using nested loops... Ok, here we go, this is my approach: :-)

import numpy as np

n = 7
for i in range(n):
    print(1-np.mod(np.arange(i+1), 2)[::-1])

Upvotes: 1

Woody1193
Woody1193

Reputation: 7970

You could try this:

def run():
    data = [1]
    iterations = 7
    for i in range(iterations):
        print_data(data)
        data.append(i % 2)
    print_data(data)

def print_data(dat):
    print(' '.join(d for d in dat[::-1]))

This solution has the added benefit of not using insert, which is O(n) and replacing an if-statement with a mathematical expression.

Upvotes: 0

rahlf23
rahlf23

Reputation: 9019

You are on the right track with your comment, here is one option using insert():

start = [1]
no_rows = 7

for i in range(no_rows):
    print(start)
    start.insert(0, 1 if start[0]==0 else 0)

Which gives:

[1]
[0, 1]
[1, 0, 1]
[0, 1, 0, 1]
[1, 0, 1, 0, 1]
[0, 1, 0, 1, 0, 1]
[1, 0, 1, 0, 1, 0, 1]

If you want each line to be formatted as a str opposed to a list, then you can change print(start) to print(' '.join([str(s) for s in start])), which gives:

1
0 1
1 0 1
0 1 0 1
1 0 1 0 1
0 1 0 1 0 1
1 0 1 0 1 0 1

As per Patrick Haugh's comment, you could instead simply replace print(start) with print(*start) to print the list of integers as a str.

Upvotes: 2

slider
slider

Reputation: 12990

You could do something like this:

s = ''
for i in range(1, 8):
    s = str(i % 2) + s
    print(s)

Upvotes: 4

Related Questions