#Without Exceptional handling program terminates if key b not found
s={'d':2,'k':4}
if b in s.keys():
s['b'].append[9]
else:
s['b']= 8
#Exceptional handling for key error and program run without termination
s={'d':2,'k':4}
try:
s['b'].append[9]
except KeyError:
s['b']= 8
Solve Problems by Coding Solutions - A Complete solution for python programming
Reverse of a number using functions
def intreverse(Number):
Reverse = 0 # Initialize the reverse value to overcome garbage value storage
while(Number > 0):
Reminder = Number %10
Reverse = (Reverse *10) + Reminder
Number = Number//10
return (Reverse)
Reverse = 0 # Initialize the reverse value to overcome garbage value storage
while(Number > 0):
Reminder = Number %10
Reverse = (Reverse *10) + Reminder
Number = Number//10
return (Reverse)
What is the type of each of the following expressions (within the type function)?
print type(5) <type 'int'>
print type("abc") <type 'str'>
print type(True) <type 'bool'>
print type(5.5) <type 'float'>
print type(12/27) <type 'int'>
print type(2.0/1) <type 'float'>
print type(12 ** 3) <type 'int'>
print type(5 == "5") <type 'bool'>
a = str((-4 + abs(-5) / 2 ** 3) + 321 -((64 / 16) % 4) ** 2)
print a
Ans: 317
print type("abc") <type 'str'>
print type(True) <type 'bool'>
print type(5.5) <type 'float'>
print type(12/27) <type 'int'>
print type(2.0/1) <type 'float'>
print type(12 ** 3) <type 'int'>
print type(5 == "5") <type 'bool'>
a = str((-4 + abs(-5) / 2 ** 3) + 321 -((64 / 16) % 4) ** 2)
print a
Ans: 317
New Dictionary Creation and updation
Transpose of a matrix (nested list) in python
row1 = [13,25,73]
row2 = [54,95,36]
row3 = [27,98,19]
matrix = [row1, row2, row3]
trmatrix = [[row[0] for row in matrix],[row[1] for row in matrix], [row[2] for row in matrix]]
print'Transpose of a matrix',trmatrix
source: http://stackoverflow.com
row2 = [54,95,36]
row3 = [27,98,19]
matrix = [row1, row2, row3]
trmatrix = [[row[0] for row in matrix],[row[1] for row in matrix], [row[2] for row in matrix]]
print'Transpose of a matrix',trmatrix
source: http://stackoverflow.com
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