How to run a function inside of an if-statement in Python

Question

I am trying to add variable information from an input to a text document. The document is in the place it should be. I have this code so far:

import time
import os
print("Welcome!")
name = input("Please enter your name: ")
print("Hello",name, "! I am going to guess your most favorite type of music.")
time.sleep(2)
print("Please, choose from one of the following: ")
listening_time = ["1 - One hour a day", "2 - About two hours per day", "3 - three to four hours per day", "4 - Most of the day"]
print(listening_time)
how_often = int(input("I find myself listening to music..."))

def add_file_1(new_1):
    f = open("music.txt", "a")
    f.write("1 Hour")

def add_file_2(new_2):
    f = open("music.txt", "a")
    f.write("2 Hours")

def add_file_3(new_3):
    f = open("music.txt", "a")
    f.write("3 - 4 Hours")

def add_file_4(new_4):
    f = open("music.txt", "a")
    f.write("Most of the day")

if how_often == str('1'):
    add_file_1(new_1)
elif how_often == str('2'):
    add_file_2(new_2)
elif how_often == str('3'):
    add_file_3(new_3)
else:
    add_file_4(new_4)

It looks like you define `how_often` as an int, then test its equality to strings — Brad Solomon, Jul 11 '18 at 00:52
I tried changing it to be a string to a string comparison and I get an error. Before (with the code above) I would get no error. It just wouldn’t do anything to the text file. — Jason White, Jul 11 '18 at 01:03
The question is: how can I change the code to do something. I get no error the way it is. It just doesn’t do anything to the text file. My last version of the code I tried putting `f = open("music.txt", "a") f.write(“2 Hours”)` inside of the if statement itself and got no result. — Jason White, Jul 11 '18 at 01:05
Consider `f.close()`. Better, use the [`with` pattern](https://stackoverflow.com/questions/6159900/correct-way-to-write-line-to-file). — Mateen Ulhaq, Jul 11 '18 at 01:08
`new_1` is the perfect example of how hard it is to name things in programming. This basically outlines the revision from putting the `f = open(“music.txt”, “a”) f.write(“2 Hours”)` inside the if statement. I changed it because I thought a function might do the trick but there was no change. — Jason White, Jul 11 '18 at 01:09

score 4 · Answer 1 · answered Jul 11 '18 at 00:54

4

You're close! You don't need to do any int-to-string conversion in your if-statement. The following will work just fine:

if how_often == 1:
    add_file_1(new_1)
elif how_often == 2:
    add_file_2(new_2)
elif how_often == 3:
    add_file_3(new_3)
else:
    add_file_4(new_4)

As Brad Solomon mentioned, the reason it's not working is because how_often is an int, but you're comparing it to a string and they are not equal.

Visit https://repl.it/repls/ScaredGhostwhiteRegister to see this codin action. While the function won't actually load, you can see which function it's trying to call based on the input you provide.

answered Jul 11 '18 at 00:54

Sheil Naik

41
4

I’m now getting an error that `add_file_2` is not defined. Although `add_file_1` got no error – Jason White Jul 11 '18 at 01:13
*Comment Update* With any number I press (1-4) I get the “Not Defined” error. – Jason White Jul 11 '18 at 01:14

score 0 · Answer 2 · answered Jul 11 '18 at 01:14

Why would you use the func?

I don't think this code will be necessary.

If I were you, I would use file openings globally.

Source :

import time
import os
print("Welcome!")
name = input("Please enter your name: ")
print("Hello",name, "! I am going to guess your most favorite type of music.")
time.sleep(2)
print("Please, choose from one of the following: ")
listening_time = ["1 - One hour a day", "2 - About two hours per day", "3 - three to four hours per day", "4 - Most of the day"]
print(listening_time)
how_often = int(input("I find myself listening to music..."))

f =open("music.txt", "a") 
if how_often == 1:
    f.write("1 Hour")
elif how_often == 2:
    f.write("2 Hours")
elif how_often == 3:
    f.write("3 - 4 Hours")
else:
    f.write("Most of the day")

If I don't understand, let me know.

This worked perfectly. Thanks for your help! – Jason White Jul 11 '18 at 01:19 — Jason White, Jul 11 '18 at 01:19

Muzol · Answer 3 · 2018-07-11T01:41:46.700

0

You will get an error because you are comparing an integer and a string in how_often == str('1'):. But you can fix that like this: how_often == 1:.

You don't need to create a function add_file for different types of input you can create just one and make it generic:

def add_file(new_1, usr_hours_choise):
    with open("music.txt", "a") as f:
        f.write(usr_hours_choise)

and then in the if statement you assign it:

if how_often == 1:
    add_file(new_1, "1 hour.")
elif how_often == 2:
    add_file(new_2, "2 hour.")
elif how_often == 3:
    add_file(new_3, "3 - 4 hour.")
elif how_often == 1:
    add_file(new_4, "most of the day.")

This will save a lot of lines of code. But it is not clear what new_1 and what other ones do.

edited Jul 11 '18 at 01:41

answered Jul 11 '18 at 01:25

Muzol

301
1
11

1

it's good practice to use 'elif' when chaining 'if' conditionals where only one condition will be true. – Todd Jul 11 '18 at 01:40
I forgot this thanks. – Muzol Jul 11 '18 at 01:41

score 0 · Answer 4 · answered Jul 11 '18 at 01:32

you could simplify things with a dictionary for the listening time options. And a good way to open files is to use a 'with' block.

print("Please, choose from one of the following: ")

listening_times = {1:"One hour a day", 
                   2:"About two hours per day", 
                   3:"Three to four hours per day", 
                   4:"Most of the day"}

for k in listening_times:
    print "    %d - %s" % (k, listening_times[k])

how_often = int(input("I find myself listening to music..."))

with open("music.txt", 'a') as f:
    f.write(listening_times[how_often])

How to run a function inside of an if-statement in Python

4 Answers4