Review: f-strings, `.format()` and Accumulators

String Formatting
- f-strings
- .format()
Accumulator Review!
String Accumulators
List Accumulators
Dictionary Accumulators

f-strings

Your go-to for string interpolation
Curly braces contain variable reference, operations, function calls, etc

var1, var2 = 2, 3

print(f'{var1} out of {var2} programmers prefer f-strings')

`.format()`

You may still see and use this, but f-strings are generally preferable.
Interpolates values into strings using the .format() method

var1, var2 = 1, 3

print('{} out of {} programmers prefer the format method'.format(var1, var2))

Accumulators

An accumulator can be thought of as a running total that is held in a variable
Accumulator Pattern:

Initialize the accumulator variable
Repeat:
    Modify the accumulator variable

When the above loop terminates, the accumulator will have the correct value.

Sum accumulators using `int` or `float`

We can calculate a sum by using an numeric typed accumulator

num = 0
for x in list_of_numbers:
	num += x

BREAKOUT (4 minutes)

Write a function called accum_nums that takes in an integer n as an argument it should perform the tasks below for a range of numbers from 1 through n (int accumulator starts at 1)

If the number is divisible by 3, add 3 to the accumulator
If the number is divisible by 5, divide accumulator by 5
If number is divisible by 4, multiply the accumulator by 4
If number is divisible by 3, 4, and 5 do nothing
If number is divisible by 3 and 4, subtract 12 from accumulator
If number is divisible by 3 and 5, floor divide accumulator by 15
If number is divisible by 4 and 5, modulo accumulator by 20

BREAKOUT SOLUTION

def accum_nums(n):
    acc = 1
    for num in range(1, n+1):
        if num % 4 == 0 and num % 5 == 0 and num % 3 == 0:
            continue
        if num % 4 == 0 and num % 3 == 0:
            acc -= 12
        if num % 3 == 0 and num % 5 == 0:
            acc //= 15
        if num % 4 == 0 and num % 5 == 0:
            acc %= 20
        if num % 3 == 0:
           acc += 3
        if num % 4 == 0:
            acc *= 4
        if num % 5 == 0:
            acc /=5
    return acc

Accumulators using a `string`

We can concatenate a string by using a string type accumulator

str1 = ''
for s in string:
    str1 += s

We can concatenate the string by letter, by word, or by some other separator within our string

BREAKOUT (4 minutes)

Write a function called build_non_vowel_str() that takes in a string and returns a string accumulator that accumulates all non-vowels.

Pass in this string:

string = "It’s a beautiful day in the neighborhood, A beautiful day for a neighbor, could you be mine? Would you be mine?"

BREAKOUT SOLUTION

def non_val(s):
	new_s = ''
	for char in s:
		if char.lower() in ['a', 'e', 'i', 'o', 'u']:
			continue
		else:
			new_s += char
    return new_s

BREAKOUT (4 minutes)

Write a function called collect_evens() that takes a list of integers. The function should return a string that accumulates the even numbers into a string.

Example: collect_evens([1, 2, 3, 4]) -> '24'

BREAKOUT SOLUTION

def collect_evens(lst):
	s = ''
	for i in l:
		if i % 2 == 0:
			s += str(i)
	return s

`list` Accumulators

Sometimes, you will need an accumulator that is an empty list.
As the for loop gets evaluated, the accumulator will be appended with new values.
Syntax:

accumulator = [ ]
for element in some_list:
	# ...some code...
	accumulator.append(some_value)

BREAKOUT (4 minutes)

Write a function called words_start_with() with two parameters, a list of letters and a string. The function should return a new list filled with words from the string that start with the letters in the list of letters. Test with this string:

str1 = "It's a beautiful day in the neighborhood, A beautiful day for a neighbor, could you be mine? Would you be mine?"

BREAKOUT (4 minutes)

def words_start_with(letters, string):
    lst = []
    for word in string.split():
        if word[0].lower() in letters:
            lst.append(word)
    return lst

BREAKOUT (4 minutes)

Write a function called examine_lst() that takes a list of various values as an argument, then returns a new list of those values processed by these rules:

Accumulate elements that are not integers
- If an element is a float, change it to a string
- If an element is a string, get the length of that string
Test String:

l1 = [192, 504, 23.11, 3.14, 'table', 'chair', 55, 1039.1, 0, 0.0, '0.0', 'python']

BREAKOUT SOLUTION

def examine_lst(l):
    out_l = []
    for i in l:
        if type(i) == float:
            out_l.append(str(i))
        elif type(i) == str:
            out_l.append(len(i))
	    elif type(i) != int:
            out_l.append(i)
    return out_l

Dictionary Accumulators

Sometimes, you will need an accumulator that is a dictionary.
The accumulator either create a new key value pair or it will add something to the value of an already existing key value pair.
Syntax for a dictionary counter:

accumulator = {}
for element in some_list:
	if element in accumulator.keys():
		accumulator[element] += 1
	else:
		accumulator[element] = 1

Dictionary Accumulators Code Along ...

Write a function that will:

For any given string, return a dictionary that gives us the number of vowels, the number of non vowels, the number of non-alpha characters.

Let’s do this together!

Dictionary Accumulators Code Along ...

Step 1: We need to first create a dictionary that we can store all our information in.

d = {'vowels': 0, 'consonants': 0, 'nonalpha': 0}

We can fill our dictionary like this because we know exactly what the question is asking and because we want counts, we can use integer accumulators for our values using addition.

Dictionary Accumulators Code Along ...

Step 2:

d = {'vowels': 0, 'consonants': 0, 'nonalpha': 0}

We now need to iterate over our string and identify each of the three types of characters we are interested in.

How can we identify vowels in a string?

Dictionary Accumulators Code Along ...

Step 2:

d = {'vowels': 0, 'consonants': 0, 'nonalpha': 0}

We now need to iterate over our string and identify each of the three types of characters we are interested in.

How can we identify vowels in a string?

if char.lower() in ['a', 'e', 'i', 'o', 'u']:
    # do something

Dictionary Accumulators Code Along ...

Step 2:

d = {'vowels': 0, 'consonants': 0, 'nonalpha': 0}

We now need to iterate over our string and identify each of the three types of characters we are interested in.

How can we identify non-alphas in a string?

Dictionary Accumulators Code Along ...

Step 2:

d = {'vowels': 0, 'consonants': 0, 'nonalpha': 0}

We now need to iterate over our string and identify each of the three types of characters we are interested in.

How can we identify non-alphas in a string?

elif not char.isalpha():
    # do something

Dictionary Accumulators Code Along ...

Step 2:

d = {'vowels': 0, 'consonants': 0, 'nonalpha': 0}

We now need to iterate over our string and identify each of the three types of characters we are interested in.

Based on this, we can use just a regular else statement to cover all other characters

Dictionary Accumulators Code Along ...

Step 2:

d = {'vowels': 0, 'consonants': 0, 'nonalpha': 0}

Let’s put it all together - you now fill in the blanks of how to use our dictionary to accumulate here

if char in ['a', 'e', 'i', 'o', 'u']:
	pass
elif not char.isalpha():
	pass
else:
	pass

Dictionary Accumulators Code Along END

Step 2:

d = {'vowels': 0, 'consonants': 0, 'nonalpha': 0}

Let’s put it all together - you now fill in the blanks of how to use our dictionary to accumulate here

if char in ['a', 'e', 'i', 'o', 'u']:
	d['vowels'] += 1
elif not char.isalpha():
	d['nonalpha'] += 1
else:
	d['consonants'] += 1

Dictionary Accumulators Code Along #2 ...

Write a function that take in an integer and will roll a 6-sided dice randomly for that number of times and will update the proportions of each number occurring and update the counts also

Have your function return:
- A dictionary of counts
- A dictionary of dice proportions

Dictionary Accumulators Code Along #2 ...