Added some mterials for the first hackathon; Updated answers in solut…

…ions for second section

Hackathon_Sessions/Session1_LoopsLogic_Python/SOLUTIONS_CnC_Loops_and_Logic.py

@@ -0,0 +1,163 @@
+# -*- coding: utf-8 -*-
+"""
+Coffee and Coding Python Exercises
+
+Session Focus:
+For Loops and If Statements 
+
+@authors: Craig Scott, Ian Banda
+"""
+
+#%%
+# Exercise 1
+
+def interger_sum(begin_val, end_val, step_val):
+    '''
+    Sums integers between specified values. 
+
+    Parameters
+    ----------
+    begin_val : int
+        value to begin sequence.
+    end_val : int
+        value to end sequence.
+    step_val : int
+        step_val the integers steps between numbers.
+
+    Returns
+    -------
+    int
+        A sum of the integers
+
+    '''
+    # Check whether the begin value is greater than end value - return 0 if True
+    if begin_val > end_val:
+
+        return 0
+
+    else:
+        # Return sum of integers 
+        return sum(range(begin_val,end_val+1,step_val))
+
+
+# test integer_sum using the following tests
+
+begin_val = 4
+end_val = 4
+step_val = 4
+
+interger_sum(begin_val, end_val, step_val)
+# should return 4
+
+interger_sum(3,11,2)
+# should return 35 (3 + 5 + 7 + 9 + 11)
+
+interger_sum(1,15,5)
+# should return 18 (1+6+11)
+
+interger_sum(9,7,2)
+# should return 0 (begin_val is larger than end_val)
+
+
+#%%
+# Exercise 2
+
+def odd_int_count(int_array):
+    '''
+    Returns the integer that appears an odd number of time. 
+
+    Parameters
+    ----------
+    int_array : list
+        list of integers.
+
+    Returns
+    -------
+    i : int
+        The integer that appears an odd number of times.
+
+    '''
+
+    # Use set to only get the unique values in the array
+    unique_int_array = set(int_array)
+
+    # cycle through the unique set of integers 
+    for i in unique_int_array:
+
+        # use modulo to get the remainder of the division of the number of times it appears divided by 2
+        # If this is 1 - the number is 1 and we return this number
+        if int_array.count(i) % 2:
+
+            return i
+
+
+# test odd_int_count using the following tests
+
+odd_int_count([2,2,2,5,5,5,5])
+# should return 2 (it appears an odd number of times)
+
+odd_int_count([1,2,2,2,3,3,1,1,1])
+# should return 2 (it appears an odd number of times)
+
+odd_int_count([10,10,10,10,3,4,5,4,2,1,1,5,5,3,2,10,5])
+# should return 10 (it appears an odd number of times)
+
+
+#%%
+# Exercise 3
+
+def move_zeros(int_array):
+    '''
+    Moves zeros in an array to the end of the list. 
+
+    Parameters
+    ----------
+    int_array : list
+        list of ints.
+
+    Returns
+    -------
+    int_array : list
+        A reordered list with zeros at the end.
+
+
+    '''
+
+    # create some values to control the while loop
+    still_zeros = 1
+    number_of_zeros = 0
+
+    # While the variable still_zeros is still true 
+    while still_zeros:
+
+        # catch an error when there are no more zeros
+        try:
+
+            # get the index of he next zero
+            zero_index = int_array.index(0)
+            # increase the number zeros counter
+            number_of_zeros += 1
+            # use the pop function to remove that zero value 
+            int_array.pop(zero_index)number.pop
+
+        except ValueError:
+            # if there are no more zeros break the while loop by changing still zeros 
+            # to be 0
+            still_zeros = 0
+
+    # for the number of times there was a zero in the list 
+    for i in range(number_of_zeros):
+        # use append to add a zero at the end 
+        int_array.append(0)
+
+    return int_array
+
+
+move_zeros([3,4,0,5,6])
+# should return [3,4,5,6,0]
+
+move_zeros([11,25,0,0,1,3,4,5,11,10,29,0,3])
+# Should return [11,25,1,3,4,5,11,1,29,3,0,0,0]
+
+move_zeros([0,5,0,8,9,0,4,9,0])
+# should return [5,8,9,4,9,0,0,0,0]