Truncate decimal places of values within a pandas df

Question

I can truncate individual floats using the truncate function in math. But when trying to pass the same function to a pandas df column I'm getting an error.

import math
import pandas as pd

X = 1.1236

X = math.trunc(1000 * X) / 1000;

#Output
1.123

But when using a pandas df:

d = ({
    'X' : [1.1234,1.1235],           
    })

df = pd.DataFrame(data=d)

df['X'] = math.trunc(1000 * df['X']) / 1000;

Error:

df['X'] = math.trunc(1000 * df['X']) / 1000;

TypeError: type Series doesn't define __trunc__ method

Comparing to other figures with _n_ decimal places. 1.123 billion is a big difference from 1.124 billion — jonboy, Jun 27 '19 at 00:33

score 8 · Accepted Answer · edited Jun 27 '19 at 00:35

8

You can use applymap

trunc = lambda x: math.trunc(1000 * x) / 1000;

df.applymap(trunc)

edited Jun 27 '19 at 00:35

Lucas

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answered Jun 27 '19 at 00:16

Anthony Kong

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score 6 · Answer 2 · answered Aug 16 '20 at 23:48

6

I believe the easiest way to achieve this would be using .astype(int) In your example, it would be:

df[x] = ((df[x]*1000).astype(int).astype(float))/1000

answered Aug 16 '20 at 23:48

Gustavo Linari Rodrigues

141
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score 2 · Answer 3 · answered Jun 27 '19 at 00:20

2

Try changing df['X'] = math.trunc(1000 * df['X']) / 1000; to df['X'] =[math.trunc(1000 * val) / 1000 for val in df['X']]. Hope it helps

answered Jun 27 '19 at 00:20

TavoGLC

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score 1 · Answer 4 · answered Sep 05 '22 at 16:30

1

A simple way is converting it into integer values, like this:

df['X'] = df['X'].astype(int)

answered Sep 05 '22 at 16:30

Haddock-san

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Truncate decimal places of values within a pandas df

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