Python removing first 3 digits in a number

Question

Hopefully a simple one, I have a number, say 1234567.890 this number could be anything but will be this length.

How do I truncate the first 3 numbers so it turns into 4567.890?

This could be any number so subtracting 123000 will not work.

---

I'm working with map data in UTM coordinates (but that should not matter)

Example

x = 580992.528
y = 4275267.719

For x, I want 992.528

For y, I want 267.719

Note: y has an extra digit to the left of the decimal so 4 need removing

Answer 1

You can use slices for this:

x = 1234567.890

# This is still a float
x = float(str(x)[3:])

print(x)

Outputs:

4567.89

---

As [3:] gets the starts the index at 3 and continues to the end of the string

Answer 2

Update after your edit

The simplest way is to use Decimal:

from decimal import Decimal

def fmod(v, m=1000, /):
    return float(Decimal(str(v)) % m)

print(fmod(x))
print(fmod(y))

Output

992.528
267.719

---

If you don't use string, you will have some problems with floating point in Python.

Demo:

n = 1234567.890
i = 0
while True:
    m = int(n // 10**i)
    if m < 1000:
        break
    i += 1
r = n % 10**i

Output:

>>> r
4567.889999999898

>>> round(r, 3)
4567.89

Same with Decimal from decimal module:

from decimal import Decimal

n = 1234567.890
n = Decimal(str(n))
i = 0
while True:
    m = int(n // 10**i)
    if m < 1000:
        break
    i += 1
r = n % 10**i

Output:

>>> r
Decimal('4567.89')

>>> float(r)
4567.89

Answer 3

This approach simply implements your idea.

• int_len is the length of the integer part that we keep
• sub is the rounded value that we will subtract the original float by

Code

Here is the code that implements your idea.

import math


def trim(n, digits):
    int_len = len(str(int(n))) - digits  # length of 4567
    sub = math.floor(n / 10 **int_len) * 10**int_len

    print(n - sub)

But as Kelly Bundy has pointed out, you can use modulo operation to avoid the complicated process of finding the subtrahend.

def trim(n, digits):
    int_len = len(str(int(n))) - digits  # length of 4567

    print(n % 10**int_len)

Output

The floating point thing is a bit cursed and you may want to take Corralien's answer as an alternative.

>>> n = 1234567.890

>>> trim(n, 3)
4567.889999999898

Answer 4

def get_slice(number, split_n):
  return number - (number // 10**split_n) * 10**split_n