CODEWITHSUNDEEP
python
Sam Heather
Sam Heather

Reputation: 1503

Python - can't convert to int

My entire code is below (didn't want to miss anything). For some reason I keep getting an error where I can't convert a float to an int?

import math

def getRange(input):
    if (1 <= input < 10):
        return [1,1,9]
    elif (10 <= input < 190):
        return [2,10,99]
    elif (190 <= input < 2890):
        return [3,100,999]
    elif (2890 <= input < 38890):
        return [4,1000,9999]
    elif (38890 <= input < 488890):
        return [5,10000,99999]
    elif (488890 <= input < 5888889):
        return [6,100000,999999]

def getDigit(input):
    workingRange=getRange(input)
    multi_digit_dec = ((input-workingRange[1])/workingRange[0])+workingRange[1]
    multi_digit_float = math.floor((input-workingRange[1])/workingRange[0])+workingRange[1]
    print multi_digit_float
    multi_digit_int = input(multi_digit_float)
    decimal_remainder = multi_digit_int - multi_digit_dec
##    digit_id = decimal_remainder * len(str(multi_digit_int))
##    actual_digit = str(multi_digit_dec)[digit_id]
##    return actual_digit


getDigit(100)

My error is:

Traceback (most recent call last):
  File "C:\Users\Samuel\Desktop\Python\concatenate string of variables and product values.py", line 29, in <module>
    getDigit(100)
  File "C:\Users\Samuel\Desktop\Python\concatenate string of variables and product values.py", line 22, in getDigit
    multi_digit_int = int(multi_digit_float)
TypeError: 'int' object is not callable
>>>

Code updated above to reflect change of variable called int to input

Upvotes: 2

Views: 580

Answers (5)

James Waldby - jwpat7
James Waldby - jwpat7

Reputation: 8701

In your stack of elif statements, if you start by testing if the input number is less than 1, then there's no need to check the lower ends of ranges other than the first. This will cut out half of the verbiage in the elif's. Also, it is more compact to use a loop for tests like this. For example, the following code produces the output shown below it.

def getRange(k):
    if k < 1: return None
    e = 1
    for d in [10, 190, 2890, 38890, 488890, 5888889]:
        if k<d:
            return [e, 10**(e-1), 10**e -1]
        e += 1
    return None

for i in range(14):
    print '{:8} {:20}'.format(i * 3**i, getRange(i * 3**i)),
    if i&1: print

Output:

       0 None                        3 [1, 1, 9]           
      18 [2, 10, 99]                81 [2, 10, 99]         
     324 [3, 100, 999]            1215 [3, 100, 999]       
    4374 [4, 1000, 9999]         15309 [4, 1000, 9999]     
   52488 [5, 10000, 99999]      177147 [5, 10000, 99999]   
  590490 [6, 100000, 999999]   1948617 [6, 100000, 999999] 
 6377292 None                 20726199 None

Upvotes: 1

martineau
martineau

Reputation: 123443

The problem with the line multi_digit_int = input(multi_digit_float) in your updated code is that the argument input in def getDigit(input): is hiding the built-in input() function. Since you calle getDigit with an argument of 100, inside the function input is an int instance so the input(multi_digit_float) part is being interpreted as 100(multi_digit_float), i.e. an int calling something.

Upvotes: 0

sampson-chen
sampson-chen

Reputation: 47267

I think you want:

multi_digit_int = math.floor(multi_digit_float)

Also, don't use int as a variable name.

Upvotes: 0

NPE
NPE

Reputation: 500257

The problem is that you're using int as a variable name, and that shadows the built-in function. Rename the variable.

In general, it's worth familiarizing oneself with the names of the built-in functions, to avoid this type of problems.

Upvotes: 6

Oleksii Kachaiev
Oleksii Kachaiev

Reputation: 6234

Don't use int as variable name (function getDigit).

Upvotes: 4

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