

In [2]:

    
word = 'lead'
print word[0]
print word[1]
print word[2]
print word[3]









    



l
e
a
d



In [3]:

    
word = 'tin'
print word[0]
print word[1]
print word[2]
print word[3]









    



t
i
n






    



---------------------------------------------------------------------------
IndexError                                Traceback (most recent call last)
<ipython-input-3-7974b6cdaf14> in <module>()
      3 print word[1]
      4 print word[2]
----> 5 print word[3]

IndexError: string index out of range



In [4]:

    
word = 'lead'
for character in word:
    print character









    



l
e
a
d



In [5]:

    
word = 'tin'
for character in word:
    print character









    



t
i
n



In [10]:

    
word = 'oxygen'
for character in word:
    print character
    character = 'Z'
print "done"









    



o
x
y
g
e
n
done



In [11]:

    
word = 'oxygen'
for char in word:
    print char









    



o
x
y
g
e
n



In [14]:

    
length = 0
for vowel in 'aeiou':
    length = length + 1
print "There are", length, 'vowels'









    



There are 5 vowels



In [15]:

    
print len('aeiou')



In [16]:

    
'Z' * 3









    Out[16]:





'ZZZ'



In [17]:

    
cube = 1
for count in 'Z' * 3:
    cube = cube * 5
print cube



In [18]:

    
print range(3)









    



[0, 1, 2]

Challenge:

Write code using for loop and range() that takes a number and computes its exponent.

E.g. if you have 2 and 3, the answer should be 8. Use print to display the result.

Solution:

Use result to accumulate the result in, so that the sum for 5^3 effectively become 1*5*5*5. Note that the loop variable isn't actually used (it is a simple counter), we the anonymous variable _ has been used as a loop variable.



In [4]:

    
number = 5
exponent = 3
result = 1
for _ in range(exponent):
    result = result * number
print number

Reverse a string

Given a string, print its reverse.



In [3]:

    
string = 'abcdef'
reversed_string = ''
length = len(string)
for count in range(length):
    reversed_string = reversed_string + string[length - count - 1]
    #print (length - count - 1), reversed_string
print reversed_string









    



5 f
4 fe
3 fed
2 fedc
1 fedcb
0 fedcba
fedcba



In [ ]: