Question

In: Computer Science

In Python Create a function called ????. The function receives a "string" that represents a year...

In Python

Create a function called ????. The function receives a "string" that represents a year (the variable with this "String" will be called uve) and a list containing "strings" representing bank accounts (call this list ????).

• Each account is represented by 8 characters. The format of each account number is "** - ** - **", where the asterisks are replaced by numeric characters.

o For example, “59-04-23”.

• The two central characters of the "string" of each account represent the year in which the account was created.

o For example, the account “59-04-23” was created in 2004.

o Assume that all years are from the year 2000 onwards.

• The year in uve is represented by four characters.

o For example, "2001"

• The function must return a list with all the accounts that were created in the year indicated in ???, with each counts without the “-“ symbols.

o For example, if the accounts are [“49-01-26”, “19-01-33”, “99-01-53”, “59-04-23”] and the year of interest is "2001", then the function should return ["490126", "190133", "990153"]; note that the accounts continue in the form of a "string".

• In addition, the function must return the percentage of accounts that were created in the year indicated by ???.

o For example, if the accounts are [“49-01-26”, “19-01-33”, “99-01-53”, “59-04-23”] and the year of interest is "2001", so the function should return 75% (three accounts were created in 2001, and there are four accounts, therefore 100 × 3/4 = 75%)

Expert Solution

Solution:

The code is in ython 3.

For percentage I typecasted as str(int(percent)) to return percentage without any decimal ponts.

If you need a different format please comment and will do the needful.

I have also added comments in the code for understanding

Code:

def pic(uve,bere):
    year=uve[2:4]
    accounts=[]
    #iterating all accounts
    for x in bere:
        #if accounts matches the year
        if(x[3:5]==year):
            #append the account and remove '-'s
            accounts.append(x.replace('-',''))
    percent=100*len(accounts)/len(bere)
    #To print percent exacyly without decimls along with '%'
    percentage=str(int(percent))+'%'
    #returning both accounts and percentage
    return accounts,percentage

#Code for testing     
accounts,percentage=pic("2001",["49-01-26", "19-01-33", "99-01-53", "59-04-23"])
print(accounts)
print(percentage)

Screenshot of sample output and code:

venereology answered 2 years ago

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