Print pattern 'FUN' using Python

Question

I need to display the following 'FUN' using patterns in Python. The problem I am having is that I need to have them all on a single line with space between each character. I do understand when I use print() function that it will move to the next line. I tried searching for an example but was not successful.

Below is my code for 'F U N' which will print out in vertical order.

#Pattern F

for row in range(5):
    for col in range(7):
        if (col==0 or col==1) or ((row==0 or row==2)):
            print("F",end="")
        else:
            print(end=" ")
    print()
print()

#Pattern U

for row in range(5):    # there are 5 rows
    for col in range(7): # 7 columns
        if ((col==0 or col==6) and row<3) or (row==3 and (col==1 or col==5)) or (row==4 and col>1 and col<5):
            print("U", end="")
        else:
            print(end=" ")
    print()

print()

# Pattern N

for row in range(5):
    for col in range(9):
        if (col==0 or col==1 or col==6 or col==7) or (row==col-1): #and (col>0 and col<5)):
            print("N",end="")
        else:
            print(end=" ")
    print()

Is this what you are looking for https://stackoverflow.com/questions/493386/how-to-print-without-newline-or-space ? — Benoît Zu, Jan 16 '18 at 08:14
Why don't you just compose your strings beforehand and then print everything at once? — Thomas Kühn, Jan 16 '18 at 08:16

score 1 · Answer 1 · answered Jan 16 '18 at 08:27

I would go for not printing the characters directly, but rather save them into strings first, using dedicated functions. Once you have these strings, they are easily combined to form words:

def pattern_f():
    return [
        ''.join([
            'F' if (col == 0 or col == 1 or row == 0 or row ==2) else ' '
            for col in range(7)]) for row in range(5)
    ]

def pattern_u():
    return [
        ''.join([
            'U' if ((col==0 or col==6) and row<3) or (row==3 and (col==1 or col==5)) or (row==4 and col>1 and col<5) else ' '
            for col in range(7)]) for row in range(5)
    ]

def pattern_n():
    return [
        ''.join([
            'N' if (col==0 or col==1 or col==6 or col==7) or (row==col-1) else ' '
            for col in range(7)]) for row in range(5)
    ]


##separate printing:
for string in pattern_f():
    print(string)
print()

for string in pattern_u():
    print(string)
print()

for string in pattern_n():
    print(string)    
print()

##combining
for f,u,n in zip(pattern_f(), pattern_u(), pattern_n()):
    print(f,u,n)

This gives the following output:

FFFFFFF
FF     
FFFFFFF
FF     
FF     

U     U
U     U
U     U
 U   U 
  UUU  

NN    N
NNN   N
NN N  N
NN  N N
NN   NN

FFFFFFF U     U NN    N
FF      U     U NNN   N
FFFFFFF U     U NN N  N
FF       U   U  NN  N N
FF        UUU   NN   NN

Hope this helps.

Thank You! This definitely helped. I can use your approach to my next homework problem. — golfnut1400, Jan 16 '18 at 13:05

Print pattern 'FUN' using Python

1 Answers1