String operations in Python

in #pythonlast year
astring = "Hello Steem!"
print(astring.upper())HELLO STEEM!

I used the markdown table generator to create the table for this.

Look how easy it is to get output by inserting a small bit of code.

astring = "Hello Steem!"
print(astring.lower())hello steem!

Fun string play - from exercise in

s = "Be cool always"
#Length should be 14
print("Length of s = %d" % len(s))

#First occurrence of "a" should be at index 8
print("The first occurrence of the letter a = %d" % s.index("a"))

#Number of a's should be 2
print("a occurs %d times" % s.count("a"))

#Slicing the string into bits
print("The first five characters are '%s'" % s[:5]) # Start to 5
print("The next five characters are '%s'" % s[5:10]) # 5 to 10
print("The thirteenth character is '%s'" % s[12]) # Just number 12
print("The characters with odd index are '%s'" %s[1::2]) #(0-based indexing)
print("The last five characters are '%s'" % s[-5:]) # 5th-from-last to end

#Convert everything to uppercase
print("String in uppercase: %s" % s.upper())

#Convert everything to lowercase
print("String in lowercase: %s" % s.lower())

#Check how a string starts
if s.startswith("Be"):
print("String starts with 'Be'. Good!")

#Check how a string ends
if s.endswith("ays!"):
print("String ends with 'ays!'. Good!")

#Split the string into three separate strings,
#each containing only a word
print("Split the words of the string: %s" % s.split(" "))


Length of s = 14
The first occurrence of the letter a = 8
a occurs 2 times
The first five characters are 'Be co'
The next five characters are 'ol al'
The thirteenth character is 'y'
The characters with odd index are 'eco las'
The last five characters are 'lways'
String in uppercase: BE COOL ALWAYS
String in lowercase: be cool always
String starts with 'Be'. Good!
Split the words of the string: ['Be', 'cool', 'always']


image source - logo and google colab screenshot