When a number as a decimal it is usually a float in Python.

If you want to remove the decimal and keep it an integer (int). You can call the int() method on it like so...

>>> int(2.0)
2

However, int rounds down so...

>>> int(2.9)
2

If you want to round to the nearest integer you can use round:

>>> round(2.9)
3.0
>>> round(2.4)
2.0

And then call int() on that:

>>> int(round(2.9))
3
>>> int(round(2.4))
2

My solution involving int abuse. int rounds towards the nearest 0. I multiply it by 10**3 to affect rounding. After using int, I divide it by 10**3 to get actual results.

It's safer, as it does work with e notation.

int(15.55555 * 10**3) / 10.0**3

Use math.floor

Updated code:

import math
Banana = 1
Apple = 2
Cookie = 5

money = input("How much money have you got? ")
if int(money) >= 1:
    print("For ", money," dollars you can get ",math.floor(int(money)/int(Banana)),"bananas")
if int(money) >= 2:
    print("Or ", math.floor(int(money)/int(Apple)), "apples")
if int(money) >= 5:
    print("Or ", math.floor(int(money)/int(Cookie))," cookies")
else:
    print("You don't have enough money for any other imported elements in the script")

A super simple solution is to use strings

x = float (str (w)[:-1])
y = float (str (w)[:-2])
z = float (str (w)[:-3])

Any of the floating point library solutions would require you dodge some rounding, and using floor/powers of 10 to pick out the decimals can get a little hairy by comparison to the above.

a= 9.1234567891235

def round_down(a, decimals):
    return round(a - 0.5 * 10**(-decimals), decimals)

round_down(a, decimals=9)
9.123456789