Related
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.
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)
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.
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().
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')
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=' ')
this is my code right now:
loop_count = 1
for i in range(mystery_int):
for x in range(1,mystery_int):
print(x*loop_count, end=" ")
print (loop_count)
loop_count+=1
this is what it is supposed to print:
1 2 3 4 5
2 4 6 8 10
3 6 9 12 15
4 8 12 16 20
5 10 15 20 25
But it prints:
1 2 3 4 1
2 4 6 8 2
3 6 9 12 3
4 8 12 16 4
5 10 15 20 5
You need to range till mystery_int + 1 because in range, second argument is exclusive. So, for example, range(1,6) gives numbers from 1 to 5.
Also, I added an empty print() which basically adds a newline to match with desired output.
Using end='\t' further aligns output properly.
loop_count = 1
mystery_int = 5
for i in range(mystery_int):
for x in range(1, mystery_int + 1):
print(x * loop_count, end='\t')
print()
loop_count += 1
the range for x should be range(1,mystery_int+1), and you also incorrectly print loop_count at the end of each line (which I replaced with the empty string, just to produce a newline).
loop_count = 1
for i in range(mystery_int):
for x in range(1,mystery_int+1):
print(x*loop_count, end=" ")
print('')
loop_count+=1
Note that the loop_count variable is not really needed. You could write the program as:
for i in range(1,mystery_int+1):
for x in range(1,mystery_int+1):
print(x*i, end=" ")
print('')
or even better as:
for i in range(1,mystery_int+1):
print(*[x*i for x in range(1,mystery_int+1)], sep=" ")
you are running on two for loops in addition to using another counter, i would recommend sticking only to the loops:
for i in range(1,mystery_int+1):
for x in range(1,mystery_int+1):
print(i*x, end=" ")
print("") # new line
I am having trouble getting this python code to work right. it is a code to display pascal's triangle using binomials. I do not know what is wrong. The code looks like this
from math import factorial
def binomial (n,k):
if k==0:
return 1
else:
return int((factorial(n)//factorial(k))*factorial(n-k))
def pascals_triangle(rows):
rows=20
for n in range (0,rows):
for k in range (0,n+1):
print(binomial(n,k))
print '\n'
This is what it keeps printing
1
1 1
1
2
1
1
12
3
1
1
144
24
4
1
1
2880
360
40
5
1
1
86400
8640
720
60
6
1
1
3628800
302400
20160
1260
and on and on. any help would be welcomed.!!
from math import factorial
def binomial (n,k):
if k==0:
return 1
else:
return int((factorial(n)//factorial(k))*factorial(n-k))
def pascals_triangle(rows):
for n in range (rows):
l = [binomial(n, k) for k in range (0,n+1)]
print l
pascals_triangle(5)
output:
[1]
[1, 1]
[1, 2, 1]
[1, 12, 3, 1]
[1, 144, 24, 4, 1]
there are many wrong things.
The first one is the way you compute the values : if building a pascal triangle, you want to use the previous line to compute the current one, and not use the binomial computation (which is expensive due to the number of multiplications).
then by default, print appends a "\n"
Correct implementation:
def print_line(x):
print (" ".join(map(str,x)))
def pascals_triangle(rows):
cur_line=[1,1]
for x in range (2,rows):
new_line=[1]
for n in range (0,len(cur_line)-1):
new_line.append(cur_line[n]+cur_line[n+1])
new_line.append(1)
print_line (new_line)
cur_line=new_line
this provides the following output
$ python pascal.py
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
Your binomial function had a small bracketing mistake in it, which was giving you incorrect output:
from math import factorial
def binomial(n, k):
if k==0:
return 1
else:
return int((factorial(n)/(factorial(k)*factorial(n-k))))
def pascals_triangle(rows, max_width):
for n in range (0,rows):
indent = (rows - n - 1) * max_width
print(' ' * indent, end='')
for k in range(0, n+1):
print("{:^{w}}".format(binomial(n, k), w = max_width*2), end='')
print()
pascals_triangle(7, 2)
With the addition of a padding parameter, the output can be made to look like this:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
I have this strange problem when following a reference, this code:
for r in range(10):
for c in range(r):
print "",
for c in range(10-r):
print c,
print ""
should print out something like this:
0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8
0 1 2 3 4 5 6 7
0 1 2 3 4 5 6
0 1 2 3 4 5
0 1 2 3 4
0 1 2 3
0 1 2
0 1
0
but Instead I am getting this:
0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8
0 1 2 3 4 5 6 7
0 1 2 3 4 5 6
0 1 2 3 4 5
0 1 2 3 4
0 1 2 3
0 1 2
0 1
0
Can anyone explain to me what is causing in indent on right side, it seems so simple but I have no clue what I can do to fix this?
You were printing the leading spaces incorrectly. You were printing empty quotes ("") which is printing only a single space. When you do print c, there is a space printed after c is printed. You should print " " instead to get the correct spacing. This is a very subtle thing to notice.
for r in range(10):
for c in range(r):
print " ", #print it here
for c in range(10-r):
print c,
print ""
Test
If you want to format it just so, it might be better to just let Python do it for you instead of counting explicit and the hidden implicit spaces. See the string formatting docs for what {:^19} means and more.
for i in range(10):
nums = ' '.join(str(x) for x in range(10 - i))
#print '{:^19}'.format(nums) # reproduces your "broken" code
print '{:>19}'.format(nums) # your desired output
Using the print function is a good alternative sometimes, as you can eliminate hidden spaces by setting the keyword argument end to an empty string:
from __future__ import print_function # must be at the top of the file.
# ...
print(x, end='')
You are simply not creating enough indent on the left side (there is no such thing as right side indent while printing).
For every new line you want to increase the indent by two spaces, because you are adding a number+whitespace on the line above. "", automatically adds one whitespace (this is why there is whitespaces between the numbers). Since you need to add two, simply add a whitespace within the quotes, like this: " ",.
The extra whitespace is filling the space of the number in the line above. The comma in "", is only filling the space between the numbers. To clarify: " ", uses the same space as c,, two characters, while "", only uses one character.
Here is your code with the small fix:
for r in range(10):
for c in range(r):
print " ", # added a whitespace here for correct indent
for c in range(10-r):
print c,
print ""
I Am using python 3.3 with IEP and i am trying to make a multiplication table that is nice an orderly. Everywhere i look online says it will be nice but it ends up just being 1 row and long where i want
1 2 3 4
2 4 6 8
3 6 9 12
the code i find is generally like this... SO whats wrong with it?
def main():
i = 1
print("-" * 50)
while i < 11:
n = 1
while n <= 10:
print("%4d" % (i * n),)
n += 1
print("")
i += 1
print("-" * 50)
main()
Because there is a line break after each print
Change 7th line to
print("%4d" % (i * n), end=" ")
The problem is right here:
print("%4d" % (i * n),)
Each print call implicitly puts a line break at the end of the output, but you can change that by providing the end keyword argument to print().
You can do something like this:
In [1]: def print_table(size):
...: for i in range(1, size+1):
...: print(''.join('{:>4d}'.format(i*j) for j in range(1, size+1)))
...:
In [2]: print_table(5)
1 2 3 4 5
2 4 6 8 10
3 6 9 12 15
4 8 12 16 20
5 10 15 20 25