How is print('\r') or print(' ') giving me the output?-CodePudding

We were asked to print the following output:

1 1 1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2 2
3 3 3 3 3 3 3 3
4 4 4 4 4 4 4
5 5 5 5 5 5
6 6 6 6 6
7 7 7 7
8 8 8
9 9 
10

I understand that it would require two loops so I tired this:

a = int(input())
i = a
f = 1
while i>0:
 for j in range(i):
      print(f,end=' ')
 f  = 1
 i -= 1
 print('\r')

With this I am getting the desired output, but as soon as I remove the last line of print('\r') the output becomes something like this:

1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 5 5 5 5 5 5 6 6       6 6 6 7 7 7 7 8 8 8 9 9 10

The desired output also comes out when I used print(' ') instead of print('\r'), I don't understand why this is happening?

Ps: I am a noob coder, starting my freshman year, so please go easy on me, if the formatting is not up to the mark, or the code looks bulky.

CodePudding user response：

Probably not helping you so much but the following code produces the expected output:

a = 10
for i, j in enumerate(range(a, 0, -1), 1):
    print(*[i] * j)

# Output:
1 1 1 1 1 1 1 1 1 1   # i=1, j=10
2 2 2 2 2 2 2 2 2     # i=2, j=9
3 3 3 3 3 3 3 3       # i=3, j=8
4 4 4 4 4 4 4         # i=4, j=7
5 5 5 5 5 5           # i=5, j=6
6 6 6 6 6             # i=6, j=5
7 7 7 7               # i=7, j=4
8 8 8                 # i=8, j=3
9 9                   # i=9, j=2
10                    # i=10, j=1

The two important parameters here are sep (when you print a list) and end as argument of print. Let's try to use it:

a = 10
for i, j in enumerate(range(a, 0, -1), 1):
    print(*[i] * j, sep='-', end='\n\n')

# Output:
1-1-1-1-1-1-1-1-1-1

2-2-2-2-2-2-2-2-2

3-3-3-3-3-3-3-3

4-4-4-4-4-4-4

5-5-5-5-5-5

6-6-6-6-6

7-7-7-7

8-8-8

9-9

10

Update

Step by step:

# i=3; j=8

>>> print([i])
[3]

>>> print([i] * j)
[3, 3, 3, 3, 3, 3, 3, 3]

# print takes an arbitrary number of positional arguments.
# So '*' unpack the list as positional arguments (like *args, **kwargs)
# Each one will be printed and separated by sep keyword (default is ' ')
>>> print(*[i] * j)

CodePudding user response：

To make it all easier and prevent errors, you can simply do this:

n = 10
for i in range(1, n   1):
    txt = str(i)   " " # Generate the characters with space between
    print(txt * (n   1 - i)) # Print the characters the inverse amount of times i.e. 1 10, 10 1

Where it generates the text which is simply the number a space, then prints it out the opposite amount of times, (11 - current number), i.e. 1 ten times, 10 one time.

CodePudding user response：

I suggest using 2 or 4 spaces for indenting. Let's take a look:

a = int(input())
i = a
f = 1
while i>0:
    for j in range(i):
        print(f,end=' ')
    f  = 1
    i -= 1
    print('\r')

Notice the print(f,end=' ') within the inner loop. the end=' ' bit is important because print() appends a trailing new line \n to every call by default. Using end=' ' instead appends a space.

Now take a look at print('\r'). It does not specify end=' ', so it does still append a newline after each call to print. The fact that you additionally print a \r is inconsequential in this case. You could also just do print().

CodePudding user response：

you can do this way :

rows = 10
b = 0

for i in range(rows, 0, -1):
    b  = 1
    for j in range(1, i   1):
    print(b, end=' ')
    print('\r')

CodePudding user response：

Code:-

A basic way to do it:-

for i in range(1,11):
    for j in range(11,i,-1):
        print(i,end=' ')
    print()

CodePudding user response：

No need for multiple loops.

for i in range(1,11):
    # concatenate number   a space repeatedly, on the same line
    # yes, there is an extra space at the end, which you won't see ;-)
    print(f"{i} " * (11-i))

output:

1 1 1 1 1 1 1 1 1 1 
2 2 2 2 2 2 2 2 2 
3 3 3 3 3 3 3 3 
4 4 4 4 4 4 4 
5 5 5 5 5 5 
6 6 6 6 6 
7 7 7 7 
8 8 8 
9 9 
10

As to what's happening with your code...

A basic Python print prints on a line, meaning that it ends with a line feed (which moves it to the next line).

So, if I take your word for it, you've done all the hard work of say the first line of 10 ones with spaces, when you are done at the following point.

#your code
 f  = 1
 i -= 1

Now, so far you've avoided that line feed by changing the end parameter to print so that it doesn't end with a newline. So you have:

1 1 1 1 1 1 1 1 1 1

And still no line feed. Great!

But if you now start printing 2 2 2 2 2 2 2 2 2 , it will just get added to... the end of the previous line, without line feed.

So to force a line feed, you *print anything you want, but without the end parameter being set, so that print now ends with the linefeed it uses by default.

Example:

#without line feed
print("1 " * 3, end=' ')
print("2 " * 2, end=' ')

output:

1 1 1  2 2

Lets try printing something, anything, without a end = ' ')

print("1 " * 3, end=' ')
#now add a line by a print that is NOT using `end = ' '`
print("!")
print("2 " * 2, end=' ')

output:

1 1 1  !
2 2

OK, so now we have a line feed after ! so you jump to the next line when printing the 2s. But you don't want to see anything after the 1s.

Simples, print anything that is invisible.

print("1 " * 3, end=' ')
#now add a line by a print, but using a non-visible character.
#or an empty string.  Tabs, spaces, etc... they will all work
print(" ")
print("2 " * 2, end=' ')

output:

1 1 1
2 2

This would also work:

print("1 " * 3, end=' ')
#we could also print a linefeed and end without one...
print("\n", end="")
print("2 " * 2, end=' ')