Python Programming Series # 2 - Installation guide, Warmup Series -...

This is the continuation of the previous tutorial of the Python programming series -
So, please refer part 1 before any further.

Installation

Before starting the tutorial, I would like to go through the simplest Python Installation goto guide.
Just download Anaconda platform from this website.

Beginner's guide for Anaconda - https://classroom.udacity.com/courses/ud1111/

This will be also useful for 'Data Science' subject as well. There are many libraries, not only for Python, but also other languages as well - Javascript, etc.

Here, the notebook is 'Jupyter' which can be accessed from the command shell as follows:

jupyter notebook   # syntax for opening jupyter notebook, and launched in the browser.

This is a browser-based editor and a kernel associated with it for compilation.

Also, here another advantage is we can make separate environments for Python2 and Python3
using the following commands in the command shell for different OS - Mac, Linux, Windows:

conda create -n py2 python=2    # installs the latest version of python 2 by default

conda create -n py3 python=3    # installs the latest version of python 3 by default

And activating & deactivating the environments using the following syntaxes: -

activate py2  # activating the python2 environment named 'py2'
deactivate py2  # deactivating the activated environment 'py2'
activate py3  # activating the python3 environment named 'py3'
deactivate py3  # deactivating the activated environment 'py3'

So, in this way, the python of whichever version can be executed on the same system.

Coding

Warmup-2 > string_times

Question:
Given a string and a non-negative int n, return a larger string that is n copies of the original string.

Check for these cases:

string_times('Hi', 2) → 'HiHi'
string_times('Hi', 3) → 'HiHiHi'
string_times('Hi', 1) → 'Hi'

Solution:

def string_times(str, n):
    if(n>0):
        print(n * str)
    else:
        return -1

string_times('Hi', 2) #→ 'HiHi'
string_times('Hi', 3) #→ 'HiHiHi'
string_times('Hi', 1) #→ 'Hi'

#----------------------------------------#

Warmup-2 > front_times

Question:
Given a string and a non-negative int n, we'll say that the front of the string is the first 3 chars, or whatever is there if the string is less than length 3. Return n copies of the front;

Check for these cases:

front_times('Chocolate', 2) → 'ChoCho'
front_times('Chocolate', 3) → 'ChoChoCho'
front_times('Abc', 3) → 'AbcAbcAbc'

Solution:

def front_times(str, n):
    if(n>0 and str != ""):
        front = str[0:3]
        print(n * front)
    else:
        return -1

front_times('Chocolate', 2) #→ 'ChoCho'
front_times('Chocolate', 3) #→ 'ChoChoCho'
front_times('Abc', 3) #→ 'AbcAbcAbc'

#----------------------------------------#

Warmup-2 > string_bits

Question:
Given a string, return a new string made of every other char starting with the first, so "Hello" yields "Hlo".

Check for these cases:

string_bits('Hello') → 'Hlo'
string_bits('Hi') → 'H'
string_bits('Heeololeo') → 'Hello'

Solution:

def string_bits(str):
    nstr = ""
    for i in range(0,len(str),2):   # e.g. len = 5, so, i = 0,2,4
        nstr += str[i]
    print(nstr)

string_bits('Hello') #→ 'Hlo'
string_bits('Hi') #→ 'H'
string_bits('Heeololeo') #→ 'Hello'

#----------------------------------------#

Warmup-2 > string_splosion

Question:
Given a non-empty string like "Code" return a string like "CCoCodCode".

Check for these cases:

string_splosion('Code') → 'CCoCodCode'
string_splosion('abc') → 'aababc'
string_splosion('ab') → 'aab'

Solution:

def string_splosion(str):
    nstr = ""
    for i in range(len(str)):
        nstr += str[:i+1]     # gives 'CCoCodCode'
        #nstr += str[0:len(str)-i]  # gives 'CodeCodCoC'
    print(nstr)

string_splosion('Code') #→ 'CCoCodCode'
string_splosion('abc') #→ 'aababc'
string_splosion('ab') #→ 'aab'

#----------------------------------------#

Warmup-2 > array_count9

Question:
Given an array of ints, return the number of 9's in the array.

Check for these cases:

array_count9([1, 2, 9]) → 1
array_count9([1, 9, 9]) → 2
array_count9([1, 9, 9, 3, 9]) → 3

Solution:

def array_count9(nums):
    count = 0
    for num in nums:
        if(num == 9):
            count = count + 1
    print(count)
    
array_count9([1, 2, 9]) #→ 1
array_count9([1, 9, 9]) #→ 2
array_count9([1, 9, 9, 3, 9]) #→ 3

#----------------------------------------#

Warmup-2 > array_front9

Question:
Given an array of ints, return True if one of the first 4 elements in the array is a 9. The array length may be less than 4.

Check for these cases:

array_front9([1, 2, 9, 3, 4]) → True
array_front9([1, 2, 3, 4, 9]) → False
array_front9([1, 2, 3, 4, 5]) → False
array_front9([1, 2, 3]) #→ False

Solution:

def array_front9(nums):
  bool = False
  end = len(nums)
  if end > 4:
    end = 4
  
  for i in range(end):
    if nums[i] == 9:
      bool = True   
      print(bool)
      
  if (bool != True):    
    print(False)
      
array_front9([1, 2, 9, 3, 4]) #→ True
#print("\n")
array_front9([1, 2, 3, 4, 9]) #→ False
#print("\n")
array_front9([1, 2, 3, 4, 5]) #→ False
#print("\n")
array_front9([1, 2, 3]) #→ False

# Remarks: here, we introduced this parameter, otherwise the program would return False along with True.

#----------------------------------------#

Warmup-2 > array123

Question:
Given an array of ints, return True if the sequence of numbers 1, 2, 3 appears in the array somewhere.

Check for these cases:

array123([1, 1, 2, 3, 1]) → True
array123([1, 1, 2, 4, 1]) → False
array123([1, 1, 2, 1, 2, 3]) → True

Solution:

def array123(nums):
    max = len(nums)
    bool = False # defined for expected output, otherwise it will print False along with True.
    for i in range(max):
      if nums[i:i+3] == [1,2,3]:
        bool = True
        print(bool)
    if bool != True:
      print(False)
    
array123([1, 1, 2, 3, 1]) #→ True
print("\n")
array123([1, 1, 2, 4, 1]) #→ False
print("\n")
array123([1, 1, 2, 1, 2, 3]) #→ True

#----------------------------------------#

Warmup-2 > string_match

Question:
Given 2 strings, a and b, return the number of the positions where they contain the same length 2 substring. So "xxcaazz" and "xxbaaz" yields 3, since the "xx", "aa", and "az" substrings appear in the same place in both strings.

Check for these cases:

string_match('xxcaazz', 'xxbaaz') → 3
string_match('abc', 'abc') → 2
string_match('abc', 'axc') → 0

Solution:

def string_match(str1, str2):
  # define the minimum length of the strings
  short = min(len(str1), len(str2))
  #define the count
  count = 0
  
  for i in range(short):
    if str1[i:i+2] == str2[i:i+2]:
      count = count + 1
      print(str1[i:i+2]) # prints the 2 substring
  
  print(count)
  
string_match('xxcaazz', 'xxbaaz') #→ 3
print("\n")
string_match('abc', 'abc') #→ 2
print("\n")
string_match('abc', 'axc') #→ 0