Conditional If statements with decimal constraints

Question

Say I have a condition that if a number ends with.00 add 2 to it and if a number ends with.99 add 3 to it. So if I had 5.00 and 5.99, I want to write a nested if statement that would allow me to add the right value to the original number based on its ending.

How to do that in R (or python)?

what do you mean by right value? so do you need 7.00 and 8.99 as your answer? — Onyambu, Aug 10 '18 at 16:01
What form is your number in? Note that if it's a float, your representation is not guaranteed to have the exact value that you expect. — Prune, Aug 10 '18 at 16:05
@onyambu, by "right value" I meant adding the corresponding value, so either the 2 or the 3 based on the ending of the original number. — Ef. O, Aug 10 '18 at 16:20

Olivier Melançon · Accepted Answer · 2018-08-10T16:14:19.180

1

Due to floating point precision, you will need to round your numbers to two decimals as values such as 5.99 cannot be represented exactly. This means a value such as 5.9921875 will have to fulfill your criterion as finishing with 0.99.

If you are fine with that, using Python's Format Specification Mini-Language will round implicitly and allow to extract the required decimals.

def get_decimals(num, n=2):
    return '{0:.2f}'.format(num)[-2:]

def func(num):
    decs = get_decimals(num)

    if decs == '00':
        return num + 2
    elif decs == '99':
        return num + 3
    else: return num

edited Aug 10 '18 at 16:14

answered Aug 10 '18 at 16:09

Olivier Melançon

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This is great, thank you. Any idea how i can do something similar in R? – Ef. O Aug 10 '18 at 16:24
@Oliver nice use of f strings – modesitt Aug 10 '18 at 16:26
@Ef.O Unfortunately I have little if any knowledge of R. – Olivier Melançon Aug 10 '18 at 16:27
@modesitt This is not actually using an f-string. I wanted to, but decided to keep it simple as it is not a feature commonly used yet. – Olivier Melançon Aug 10 '18 at 16:27
No problem will figure it out. Thank you so much for your response, it's really helpful. - @OlivierMelançon – Ef. O Aug 10 '18 at 16:32

modesitt · Answer 2 · 2018-08-10T16:28:05.250

0

In python.

def func(num):
    if round(num, 2) - int(num) < 0.001:
        return num + 2
    elif round(num, 2) - int(num) > 0.989:
        return num + 3 
    return num

You can round the number to 2 decimal places, subtract it's integer component, then round the result to check if it is close to 0.00 or 0.99. Simply checking if the result before rounding is 0.00 or 0.99 is unreliable because of floating point numbers.

edited Aug 10 '18 at 16:28

answered Aug 10 '18 at 16:02

modesitt

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Wouldn't you want to check against 0.005 instead of 0.001 and similarly for the second condition? – Olivier Melançon Aug 10 '18 at 16:16
I think because the float is reasonable up to 12 digits, these boundaries are tighter but sufficient @OlivierMelançon – modesitt Aug 10 '18 at 16:18
I toyed with your code and probably misread it at first because it actually works like a charm. Forget my previous comment – Olivier Melançon Aug 10 '18 at 16:26

Onyambu · Answer 3 · 2018-08-10T17:11:59.363

0

It seems there are other values in your list/vector you need to deal with. I will give a vectorized format:

in R:

data = c(5.00,5.99,5.61)
mm = function(x){
  x = round(x,2)
  y = x-as.integer(x)
  ifelse(round(y-0.00,2)==0, x+2,ifelse(round(y-0.99,2)==0,x+3,x))
} 

mm(data)
[1] 7.00 8.99 5.61

In Python: vectorized format with numpy module

import numpy as np
x = np.array([5.00,5.99,5.61])
def mm(x):
    x = np.round(x,2)
    y = x - np.array(x,dtype = 'i')
    return np.where(np.round(y-0.00,2)==0, x+2,np.where(np.round(y-0.99,2)==0,x+3,x))

mm(x)
array([7.  , 8.99, 5.61])

edited Aug 10 '18 at 17:11

answered Aug 10 '18 at 16:04

Onyambu

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I think Op is implying that the number should not be changed if the the number was 5.98 for example. my original solution did this too. – modesitt Aug 10 '18 at 16:07
@modesitt what exactly do you mean? – Onyambu Aug 10 '18 at 16:12

akrun · Answer 4 · 2018-08-10T16:29:14.030

0

In R we can do

v1 + (floor(v1) != v1) + 2
#[1] 7.00 8.99

If we are looking for last two numeric elements

v2 <- sub(".*(.{2})$", "\\1", sprintf('%0.2f', v1))
ifelse(v2 == "00", v1 + 2, ifelse(v2 == "99", v1 + 3, v1))
#[1] 7.00 8.99

data

v1 <- c(5, 5.99)

edited Aug 10 '18 at 16:29

answered Aug 10 '18 at 16:08

akrun

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Conditional If statements with decimal constraints

4 Answers4

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