Describing Report Tools Functions

Modify your program from Learning Journal Unit 7 to read dictionary items from a file and write the inverted dictionary to a fil

Bond [772]

Answer:

Following are the code to this question:

import ast as a #import package ast

d1= {}#defining an empty dictionary

def invert_dictionary(d1):#defining a method invert_dictionary

i = dict()#defining variable i that hold dict method

for k in d1:#defining for loo to check dictionary value

val = d1[k]#defining val variable to holds dictionary key values

for val1 in val:#defining another for loop to hold value of dictionary and inverse it

if val1 not in i:#defining if block to check value

i[val1] = [k]#hold key value

else:#defining else block

i[val1].append(k)#add value in dictionary

return i#return inverse value

with open("dict_items.txt", "r") as data:#use open method to add a list as a file

dict_item = a.literal_eval(data.read())#defining a variable that covert item as a text string in the input file

'''covert input string into dictionary item'''

e_val = input("Please enter a value")#defining variable e_val to input value

dict_item['4']=str(e_val)#holding value in dictionary

dict_item['5']='e'#holding value in dictionary

dict_item['6']='f'#holding value in dictionary

'''Invert the dictionary value '''

invert_data =invert_dictionary(dict_item)#defining variable to holding method value

'''calculating the format of each item of inverted dictionary as a text string output file.'''

f_data = {i:str(j[0]) for i,j in inverted_data.items() }#defining f_data to hold coverted value

import json as j#import package

with open('out_put.txt', 'w') as file:#use open method to open file

file.write(j.dumps(f_data))#add value

Explanation:

In the above-given code has four parts which can be defined as follows:

In the first part, a method invert_dictionary is defined that accepts a dictionary, and defines two for loops in which the second loop uses a conditional statement to convert the inverse form and return its value, and after inverse it use the open method to add a list as a file, and defining a variable that covert item as a text string in the input file.

In the second step, a variable e_val is declared for input value , and use the dict_item variable to holding dictionary value, and call the method and hold its value.

In the third and fourth step it calculating the format of each item of inverted dictionary as a text string output file.

6 0

3 years ago

Mrs. Sims polled her students about which social media app they use most. Which type of chart would best display this data?

Troyanec [42]

The best way to display this data is c. Pie chart

8 0

3 years ago

Read 2 more answers

• Open your Netbeans IDE and answer the following question

VARVARA [1.3K]

Answer:

public static void main(String[] args)

{

int cdCount;

double cdCountAfterDiscount;

DecimalFormat df = new DecimalFormat("$##.##"); // Create a Decimal Formatter for price after discount

System.out.println("Enter the amount of CD's bought");

Scanner cdInput = new Scanner(System.in); // Create a Scanner object

cdCount = Integer.parseInt(cdInput.nextLine()); // Read user input

System.out.println("CD Count is: " + cdCount); // Output user input

if(cdCount <= 14 )

{

System.out.println("There is no discount");

System.out.println("The final price is " + cdCount*3.5);

}

if(cdCount >= 15 && cdCount<=50)

{

System.out.println("You have a 1% discount.");

cdCountAfterDiscount = (cdCount *3.5)-(3.5*.01);

System.out.println("The final price is " + df.format(cdCountAfterDiscount));

}

if(cdCount >= 51 && cdCount<120)

{

System.out.println("You have a 5% discount.");

cdCountAfterDiscount = (cdCount *3.5)-(3.5*.05);

System.out.println("The final price is " + df.format(cdCountAfterDiscount));

}

if(cdCount >= 120)

{

System.out.println("You have a 10% discount.");

cdCountAfterDiscount = (cdCount *3.5)-(3.5*.1);

System.out.println("The final price is " + df.format(cdCountAfterDiscount));

}

8 0

3 years ago

A person gets 13 cards of a deck. Let us call for simplicity the types of cards by 1,2,3,4. In How many ways can we choose 13 ca

natima [27]

Answer:

There are 5,598,527,220 ways to choose 5 cards of type 1, 4 cards of type 2, 2 cards of type 3 and 2 cards of type 4 from a set of 13 cards.

Explanation:

The crucial point of this problem is to understand the possible ways of choosing any type of card from the 13-card deck.

This is a problem of combination since the order of choosing them does not matter here, that is, the important fact is the number of cards of type 1, 2, 3 or 4 we can get, no matter the order that they appear after choosing them.

So, the question for each type of card that we need to answer here is, how many ways are there of choosing 5 cards of type 1, 4 cards of type 2, 2 cards of type 3 and 2 are of type 4 from the deck of 13 cards?

The mathematical formula for combinations is $\\ \frac{n!}{(n-k)!k!}$ , where n is the total of elements available and k is the size of a selection of k elements from which we can choose from the total n.

Then,

Choosing 5 cards of type 1 from a 13-card deck:

$\frac{n!}{(n-k)!k!} = \frac{13!}{(13-5)!5!} = \frac{13*12*11*10*9*8!}{8!*5!} = \frac{13*12*11*10*9}{5*4*3*2*1} = 1,287$ , since $\\ \frac{8!}{8!} = 1$ .

Choosing 4 cards of type 2 from a 13-card deck:

$\\ \frac{n!}{(n-k)!k!} = \frac{13!}{(13-4)!4!} = \frac{13*12*11*10*9!}{9!4!} = \frac{13*12*11*10}{4!}= 715$ , since $\\ \frac{9!}{9!} = 1$ .

Choosing 2 cards of type 3 from a 13-card deck:

$\\ \frac{n!}{(n-k)!k!} =\frac{13!}{(13-2)!2!} = \frac{13*12*11!}{11!2!} = \frac{13*12}{2!} = 78$ , since $\\ \frac{11!}{11!}=1$ .

Choosing 2 cards of type 4 from a 13-card deck:

It is the same answer of the previous result, since

$\\ \frac{n!}{(n-k)!k!} = \frac{13!}{(13-2)!2!} = 78$ .

We still need to make use of the Multiplication Principle to get the final result, that is, the ways of having 5 cards of type 1, 4 cards of type 2, 2 cards of type 3 and 2 cards of type 4 is the multiplication of each case already obtained.

So, the answer about how many ways can we choose 13 cards so that there are 5 of type 1, there are 4 of type 2, there are 2 of type 3 and there are 2 of type 4 is:

1287 * 715 * 78 * 78 = 5,598,527,220 ways of doing that (or almost 6 thousand million ways).

In other words, there are 1287 ways of choosing 5 cards of type 1 from a set of 13 cards, 715 ways of choosing 4 cards of type 2 from a set of 13 cards and 78 ways of choosing 2 cards of type 3 and 2 cards of type 4, respectively, but having all these events at once is the multiplication of all them.

5 0

4 years ago

Which of the following items is not found on a web page?

navik [9.2K]

The answer is B. book mark definitely

7 0

3 years ago