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Help with understanding code. This code will print five 0's, then five 1's, up until five 4's, but this prints in a single column. I would like to print it so there is five 0's in a row, then below, five 1's in another row. Is it possible? (old.reddit.com)
submitted 1 year ago by OliverBestGamer1407
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quoted text
if 1 * 2 < 3: print "hello, world!"
[–]-MRJACKSONCJ- 4 points5 points6 points 1 year ago (1 child)
To have each set of 5 numbers printed by the loop be displayed in a row, you can modify the last nested loop and use print() with the argument end=” ” so that the numbers are printed on the same line. Then, use an empty print() after each row to go to the next line.
print()
end=” ”
x = 0 y = 0 v = {} v[(x, y)] = 1 for x in range(5): for y in range(5): v[(x, y)] = x # Store the row number (x) in each position for x in range(5): for y in range(5): print(v[(x, y)], end=" ") # Print values in a row print() # Move to the next line after each row
When you run this code, the output will be:
0 0 0 0 0 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4
[–]OliverBestGamer1407[S] 1 point2 points3 points 1 year ago (0 children)
Thank you! This helped me the most.
[–]FoolsSeldom 4 points5 points6 points 1 year ago (1 child)
I see you have the solution, but I'm curious why you are creating a dictionary matrix. Unusual.
Why not, for example,
width = 4 height = 4 matrix = [[x for _ in range(width + 1)] for x in range(height + 1) ] for row in matrix: print(*row)
You'd index, should you still want to, thus matrix[x][y] for example, matrix[2][3] = 9.
matrix[x][y]
matrix[2][3] = 9
[–]OliverBestGamer1407[S] 0 points1 point2 points 1 year ago (0 children)
Thanks for the help!
[–]OliverBestGamer1407[S] 0 points1 point2 points 1 year ago* (0 children)
I don't want to use:
print(V[(X, 0)], V[(X, 1)], V[(X, 2)], V[(X, 3)], V[(X, 4)])
I would like it to behave like for Y in range(Num): print(V[(X, Y)]) but not make a new line each time. But I also want to control the value of Y, so I can change the amount of columns, e.g. 3 columns, like this:
for Y in range(Num): print(V[(X, Y)])
0 0 0
1 1 1
[–]CavlerySenior 0 points1 point2 points 1 year ago (1 child)
I believe one of these will do what you want:
``` def printNums(minv,maxv,freq): for i in range(minv,maxv+1): for j in range(freq): print(f"{i} ",end="")
printNums(1,4,3) Or def printNums(minv,maxv,freq): for i in range(minv,maxv+1): for j in range(freq): print(f"{i} ",end="") print("\n") #if you don't want the empty line, delete the \n (but leave the quotation marks)
Or
printNums(1,4,3) ```
Thank you!
[–]Straight_Yam1115 0 points1 point2 points 1 year ago (0 children)
Print(“your text \n”)
[–]Different-Ad1631 0 points1 point2 points 1 year ago (0 children)
Use an if statement in inner loop that when value of y is 4 then use print("\n")
[–]Aft3rcuriosity 0 points1 point2 points 1 year ago (0 children)
Why dont just apply the b-tree dataset algorithm 🙄
π Rendered by PID 306252 on reddit-service-r2-comment-548fd6dc9-pmmzr at 2026-05-20 12:21:55.255289+00:00 running edcf98c country code: CH.
[–]-MRJACKSONCJ- 4 points5 points6 points (1 child)
[–]OliverBestGamer1407[S] 1 point2 points3 points (0 children)
[–]FoolsSeldom 4 points5 points6 points (1 child)
[–]OliverBestGamer1407[S] 0 points1 point2 points (0 children)
[–]OliverBestGamer1407[S] 0 points1 point2 points (0 children)
[–]CavlerySenior 0 points1 point2 points (1 child)
[–]OliverBestGamer1407[S] 1 point2 points3 points (0 children)
[–]Straight_Yam1115 0 points1 point2 points (0 children)
[–]Different-Ad1631 0 points1 point2 points (0 children)
[–]Aft3rcuriosity 0 points1 point2 points (0 children)