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Pattern printing with odd count of stars

How to print this pattern? I have done regular patterns but I am unable to get odd stars.

You could use a simple list comprehension:
n = 5
print("\n".join((a := [("* "*i).center(n*2 + 1) for i in range(1, n + 1) if i&1]) + a[::-1][1:]))
*
* * *
* * * * *
* * *
*
You can change n to anything you want

Trying to create an x pattern with stars

def x_pattern(n):
for i in range(n):
for j in range(n):
if (j == i) or (j == n-1-i):
print('*', end='')
i=i+1
j=j-1
else:
print('', end='')
print()
When I run this function it is not showing what I am trying to achieve.
x_pattern(5)
x_pattern(6)
x_pattern(7)
x_pattern(8)
Could anybody please point out what I am doing wrong?
Edit: Removed i = i+1 and j = j+1 as patrick suggested

If you get an even number, you probably only want to print one of the two middle rows:
def x_pattern(n):
skip_middle = n%2 == 0
for i in range(n):
if i == n/2 and skip_middle: # skip one of the two middle lines of '...**...'
continue
for j in range(n):
if (j == i) or (j == n-1-i):
print('*', end='')
else:
print(' ', end='') # print a space here
print()
for l in range(5,9):
x_pattern(l)
print()
Output:
* *
* *
*
* *
* *
* *
* *
**
* *
* *
* *
* *
* *
*
* *
* *
* *
* *
* *
* *
**
* *
* *
* *
Problem here is that you get one row less then you specified - thats why I suggested adding the wanted output to your question.

how to reverse the for loop in python and print the elements from left to right

This is my code which i wrote in python
n = 5
row = 2 * n - 2
for i in range(n,-1,-1):
for j in range(row):
print(end="")
row -=2
for k in range(i+1):
print('*',end=" ")
print()
The output what is get is this
* * * * *
* * * *
* * *
* *
*
i want to print this start from left to right order for example
The expected output is :-
* * * * *
* * * *
* * *
* *
*
if it's any possible way to print the elements from left to right because in most of my program i need that logic i'm searching for it please help me and even i used reversed function for loop it will reverse the loop but i'm not getting what i expect

n = 5
print(*[' '.join(' '*i + '*'*(n-i)) for i in range(n)], sep='\n')
Output:
* * * * *
* * * *
* * *
* *
*
Explanation:
for i in range(n):
chars = ' '*i + '*'*(n-i) # creating list of (i) spaces followed
# by (n-i) stars to finish a line of n elements
print(' '.join(chars)) # join prepared values with spaces

Here is a simple solution not using list comprehension:
n = 5
for i in range(n+1):
for j in range(i):
print(" ", end="")
for j in range(i+1, n+1):
print("* ", end="")
print()
Output:
* * * * *
* * * *
* * *
* *
*

My solution - I'm new to python as well:
n = 5
c = 0
for i in range(n, 0, -1):
print(" " * c + "*" * i)
c += 1
or
n = 5
c = 0
while n >= 0:
print(" " * c + "*" * n)
n -= 1
c += 1

The problem is that print itself cannot print right-justified text. But you can print a right-justified string instead of using multiple calls to print.
Here's your original code, using join instead of an inner loop:
n = 5
row = 2 * n - 2
for i in range(n,-1,-1):
for j in range(row):
print(end="")
row -=2
row_str = " ".join(["*"] * i)
print(row_str)
And here's the modified code to make the output right-justified:
n = 5
row = 2 * n - 2
whole = row + 1
for i in range(n,-1,-1):
for j in range(row):
print(end="")
row -=2
row_str = " ".join(["*"] * i).rjust(whole)
print(row_str)

to print a triangle of hash '#' and stars '*' in python

I m trying to print the following triangle in python, but not able to
*#####
**####
***###
****##
*****#
I am only able to print '*', but not hashes.
here's my code to print '*'
max = 6
for x in range(1,max+1):
for y in range(1,x+1):
print '*',
print

length = 6
for x in range(1, length+1):
print(x * '*' + (length - x) * '#') # integer * (string), concatenate the string integer times

How to Print * in a triangle shape in python?
*
* *
* * *
* * * *
* * * * *
This is one of the coding interview questions.
def print_triangle(n):
for i, j in zip(range(1,n+1), range(n,0,-1)):
print(" "*j, '* '*i, " "*j)
print_triangle(5)

Hollow Diamond in python

My goal is to create a hollow diamond using python.
Sample input:
Input an odd Integer:
9
Sample output:
*
* *
* *
* *
* *
* *
* *
* *
*
But so far, I have the following code that is not working right. Please help me to modify the code to achieve the goal above:
a=int(input("Input an odd integer: "))
k=1
c=1
r=a
while k<=r:
while c<=r:
print "*"
c+=1
r-=1
c=1
while c<=2*k-1:
print "*"
c+=1
print "\n"
k+=1
r=1
k=1
c=1
while k<=a-1:
while c<=r:
print " "
c+=1
r+=1
c=1
while c<= 2*(a-k)-1:
print ("*")
c+=1
print "\n"
k+=1
The code above return a result that is very far from my goal.
Input an odd integer: 7
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
I am actually converting the code from this post: http://www.programmingsimplified.com/c/source-code/c-program-print-diamond-pattern written in C language and will modify later for the hollow one but I can't get it... There is something wrong with my conversion..

A Hollow diamond is the solution to the equation
|x|+|y| = N
on an integer grid. So Hollow diamond as a 1-liner:
In [22]: N = 9//2; print('\n'.join([''.join([('*' if abs(x)+abs(y) == N else ' ') for x in range(-N, N+1)]) for y in range(-N, N+1)]))
*
* *
* *
* *
* *
* *
* *
* *
*

Your problem is that you keep using print. The print statement (and the function in Python 3) will add a line-break after what you printed, unless you explicitely tell it not to. You can do that in Python 2 like this:
print '*', # note the trailing comma
Or in Python 3 (with the print function) like this:
print('*', end='')
My solution
I took my own take at the problem and came up with this solution:
# The diamond size
l = 9
# Initialize first row; this will create a list with a
# single element, the first row containing a single star
rows = ['*']
# Add half of the rows; we loop over the odd numbers from
# 1 to l, and then append a star followed by `i` spaces and
# again a star. Note that range will not include `l` itself.
for i in range(1, l, 2):
rows.append('*' + ' ' * i + '*')
# Mirror the rows and append; we get all but the last row
# (the middle row) from the list, and inverse it (using
# `[::-1]`) and add that to the original list. Now we have
# all the rows we need. Print it to see what's inside.
rows += rows[:-1][::-1]
# center-align each row, and join them
# We first define a function that does nothing else than
# centering whatever it gets to `l` characters. This will
# add the spaces we need around the stars
align = lambda x: ('{:^%s}' % l).format(x)
# And then we apply that function to all rows using `map`
# and then join the rows by a line break.
diamond = '\n'.join(map(align, rows))
# and print
print(diamond)

def diamond(n, c='*'):
for i in range(n):
spc = i * 2 - 1
if spc >= n - 1:
spc = n - spc % n - 4
if spc < 1:
print(c.center(n))
else:
print((c + spc * ' ' + c).center(n))
if __name__ == '__main__':
diamond(int(input("Input an odd integer: ")))

this is not pretty, but its a function that does what you want:
def make_diamond(size):
if not size%2:
raise ValueError('odd number required')
r = [' ' * space + '*' + ' ' * (size-2-(space*2)) + '*' + ' ' * space for space in xrange((size-1)/2)]
r.append(' ' * ((size-1)/2) + '*' + ' ' * ((size-1)/2))
return '\n'.join(r[-1:0:-1] + r)
first i check to make sure its an odd number,
then i create a list of the lines from the center downwards.
then i create the last point.
then i return them as as a string, with a mirror of the bottom on top without the center line.
output:
>>> print make_diamond(5)
*
* *
* *
* *
*
>>> print make_diamond(9)
*
* *
* *
* *
* *
* *
* *
* *
*

Made it in one single loop ;)
x = input("enter no of odd lines : ") #e.g. x=5
i = int(math.fabs(x/2)) # i=2 (loop for spaces before first star)
j = int(x-2) #j=3 # y & j carry loops for spaces in between
y = int(x-3) #y=2
print ( " " * i + "*" )
i = i-1 #i=1
while math.fabs(i) <= math.fabs((x/2)-1): # i <= 1
b = int(j- math.fabs(y)) # b = (1, 3, 1) no of spaces in between stars
a = int(math.fabs(i)) # a = (1, 0, 1) no of spaces before first star
print (" "* a + "*" + " "*b + "*")
y = y-2 # 2,0,-2
i = i-1 # 1,0,-1, -2 (loop wont run for -2)
i = int(math.fabs(i)) # i = -2
print ( " " * i + "*")

The previous answer has got two corrections, which've been done here :
import math
x = int(input("enter no of odd lines : ")) #e.g. x=5
i = int(math.fabs(x/2)) # i=2 (loop for spaces before first star)
j = int(x-2) #j=3 # y & j carry loops for spaces in between
y = int(x-3) #y=2
print ( " " * i + "*" )
i = i-1 #i=1
while math.fabs(i) <= math.fabs((x/2)-1): # i <= 1
b = int(j- math.fabs(y)) # b = (1, 3, 1) no of spaces in between stars
a = int(math.fabs(i)) # a = (1, 0, 1) no of spaces before first star
print (" "* a + "*" + " "*b + "*")
y = y-2 # 2,0,-2
i = i-1 # 1,0,-1, -2 (loop wont run for -2)
i = int(math.fabs(i)) # i = -2
print ( " " * i + "*")
Note : Now this works for both python 2.5x and python 3.x
If you wish to know the difference in the two answers, then google it!

sizeChoice = 13
N = sizeChoice//2
for column in range (-N, N + 1):
for row in range (-N, N + 1):
if abs(column) + abs(row) == N:
print ("*", end = " ")
else:
print(" ", end = " ")
print ( )

this code works perfectly to print a hollow diamond where you can set diagonal length upto user requirement
n=int(input('Enter Odd Diagonal length :'))-1
j=n-1
print(' '*(n)+'*')
for i in range(1, 2*n):
if i>n:
print(' '*(i-n)+'*'+' '*(2*j-1)+'*')
j-=1
else:
print(' '*(n-i)+'*'+' '*(2*i-1)+'*')
if n>1:
print(' '*n+'*')

Hallow Diamond in Python using only print function
for i in range(1,6):
print(' '*(5-i) + '*'*1 + ' '*(i-1) + ' '*max((i-2),0) + '*'*(1%i) )
for j in range(4,0,-1):
print(' '*(5-j) + '*'*1 + ' '*(j-1) + ' '*max((j-2),0) + '*'*(1%j) )