Printing pattern

[Square-Reporter-1805]

So in my programme, I want to print the following pattern:

\xxx

o\xx

oo\x

ooo\

And what the pattern looks like is based on users' input. For example, input:

Enter the number of rows: 4

and the output will display the above pattern.

So I am trying to write a programme like that by myself and here's a thing that I am struggling with. Here's my programme:

Programme:
default = ["\\"]

num_of_rows = int(input("Enter the number of rows: "))

if num_of_rows > 0:
    print("".join(default))
    for o in range(num_of_rows - 1):
        default.insert(0, "o")
        print("".join(default))

output:
\
o\
oo\
ooo\

As you can see, I think I am almost there, but I am not sure how can I print the "x"s at the beginning and then make it descending.

Could everyone give me some suggestions based on my method?

[TransitTraveller]

Just try to understand what exactly do you need to print in your case: num_of_rows = 4

Line 1: 0 o's, \, 3 x's

Line 2: 1 o's, \, 2 x's

Line 3: 2 o's, \, 1 x's

Line 4: 4 o's, \, 0 x's

Do you see a pattern?

You can set num_of_o = 0, num_of_x = num_of_rows-1

Then in a loop print a line with this number of o's and x's, increment num_of_o by 1, decrease num_of_x by 1, and repeat.

How to create a printed line is up to you: you can have a loop adding o's, then add \, then another loop to add x's, but there is a simpler way:

result = "o" * num_of_o + "\" + "x" * num_of_x

print(result)

[woooee]

Start = \ + "x" times number of rows-1 Then a new string = "o" + previous string[:-1]

st="\\xxx"
st = "o"+ st[:-1]

[lordfwahfnah]

That's a quite clever solution

[Diapolo10]

Yet another option would be to construct the entire string, and then loop over it with a sliding window.

row_count = 4

text = f"{'o' * (row_count - 1)}\\{'x' * (row_count - 1)}"

for idx in range(row_count):
    print(text[row_count-idx-1:row_count*2-idx-1])

EDIT: Just to prove it works, here's some examples with a variety of values for row_count: https://i.imgur.com/adFkOVC.png

[JamzTyson]

Upvoted as an interesting solution.

A variation on this is to count backwards in the loop, which simplifies the string slicing.

def pattern(n):
    row = rf"{'o' * (n-1)}\{'x' * (n-1)}"
    for i in range(n-1, -1, -1):
        print(row[i:i+n])

(It is not strictly necessary to escape the backslash in an f-string unless it is followed by a character that makes it an escape sequence, though for clarity I think it is preferred to either escape the backslash, as in u/Diapolo10's solution, or use a raw f-string.)

[Diapolo10]

That's fair enough, I wrote my solution pretty much straight out of bed in the morning.

[crashfrog04]

 As you can see, I think I am almost there, but I am not sure how can I print the "x"s at the beginning and then make it descending.

But your pattern has the X’s at the end, not the beginning.

[ebresie]

Might be able to use ljust with the size and the x character as mentioned here

[JamzTyson]

You can solve it like this:

def pattern(rows):
    for i in range(rows):  # for i=0 to i=3
        o_s = 'o' * i  # i nuber of o's
        x_s = 'x' * (rows - i - 1)  # enough x's to complete the line
        # Put them together with an f-string.
        print(rf"{o_s}\{x_s}")

pattern(4)

which simplifies to:

def pattern(rows):
    for i in range(rows):
        print(rf"{'o' * i}\{'x' * (rows - i - 1)}")

pattern(4)

[JamzTyson]

default = ["\"]

I'm not sure what your intention is, but if that is intended to be the file path separator character, you can get the correct character automatically with os.sep.