PYTHON

pattern in python

rows = 6                        #Pattern(1,121,12321,1234321,123454321)
for i in range(1, rows + 1):
    for j in range(1, i - 1):
        print(j, end=" ")
    for j in range(i - 1, 0, -1):
        print(j, end=" ")
    print()

pattern program in python

k=int(input('value of a: '))
for a in range(1,k):
    for b in range(1,k):
        if a==b:
            print('A',end=" ")
        else:
            print('B',end=" ")
        
    print()
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pattern in python

j="ramkumar"
for i in range(len(j)):
    for g in range(0,i+1):
        print(j[g],end=' ')
    print('')
# end='' meaning (print the next answer , near the previous answer in same line)
#answer 1 2 3 4 5 6 like this in same row or same line(in horizontal)



name="ramkumar"
for i in range(len(name)) :
    print(name[0:i+1])

* pattern by python

def pattern(n):    
  for i in range(0,n):       
    for j in range(0, i+1): 
      print("* " , end="")         
      print("
") pattern(5

* pattern by python

v = 1
lines = 4
for i in range(lines):
    for j in range(i):
        print(v, end=' ')
        v += 1
    print( )

* pattern program in python

numbers = 10
for rowIndex in range(numbers):
    rowString = ''
    for columnIndex in range(numbers): #Same number of points
        #each row is equivalent to the previous row + 1 - use existing iterator
        rowString += str(columnIndex + rowIndex) + '	'
    print(rowString)

* pattern program in python

def pattern(n):     for i in range(0,n):         for j in range(0, i+1):              print("* " , end="")         print("
") pattern(5)

python pattern

        python
        
            
        
     import re
v = "aeiou"
c = "qwrtypsdfghjklzxcvbnm"
print(*re.findall("(?=[%s]([%s]{2,})[%s])"%(c,v,c),input(), re.I) or [-1], sep="
")