dataframe of type List to display trailing zeros in R

Question

I have a nasty problem that bugs me a lot.

I have a list (dataframe) that looks like this:

    a            b           c
1   1.00234    1.05667    1.00198

I want to round the numbers of this dataframe to two decimal number. But the trailing zeros have to be kept like the following:

    a            b           c
1   1.00         1.06       1.00

I tried the round and printf() and so on, it doesn't work, because my data is a list. It can't be coerced. However, I'd like to keep my data structure.

Anyone of you know how to solve this? I appreciate it very much!!!

Only one dataframe. data.frame(c(1.00234, 1.05667,1.00198)) like this one. How can you show it as I pointed out above? — purpleblau
– purpleblau, Commented Mar 27, 2014 at 21:09
So a single-column data frame? Your post displays the data as three columns... — Rich Scriven
– Rich Scriven, Commented Mar 27, 2014 at 21:14
sorry, it's three columns as it showed above. But that shouldn't be the problem right? You can transpose it if you want. — purpleblau
– purpleblau, Commented Mar 27, 2014 at 21:18

Josh O'Brien · Accepted Answer · 2014-03-27 21:22:12Z

3

Here you go:

df <- data.frame(a=1.00234, b=1.0567, c=1.00198, d=99999, e=.00001)

format(round(df, 2), nsmall=2)
#      a    b    c        d    e
# 1 1.00 1.06 1.00 99999.00 0.00

answered Mar 27, 2014 at 21:22

Josh O'Brien

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2 Comments

Josh O'Brien Over a year ago

Cool. If any of your columns are non-numeric, you'll need to do something more like this: data.frame(lapply(df, function(X) if(is.numeric(X)) format(round(X,2), nsmall=2) else X))

purpleblau Over a year ago

really great explanation. This is too advanced. I need to look into it. awesome man! appreciate it!

Rich Scriven · Accepted Answer · 2014-03-27 21:29:09Z

1

You can use sapply to effectively loop the rounding over the columns. I'm having trouble determining exactly what your data looks like. But if it's a data.frame with numerous rows and columns, see this example. You'll need to convert back to class data.frame if that matters, as this method of sapply will coerce the data frame to a matrix.

> dd <- data.frame(a = runif(6), b = runif(6), c = rnorm(6))
> dd
          a         b          c
1 0.3992252 0.9905755 -0.2557345
2 0.5052276 0.7990887 -0.7557547
3 0.3215714 0.1134675 -0.4389722
4 0.1794793 0.5372685  1.1657751
5 0.9543305 0.8908360 -1.5966621
6 0.9525730 0.5991279 -0.4819168

> as.data.frame(sapply(dd, round, 2))
     a    b     c
1 0.40 0.99 -0.26
2 0.51 0.80 -0.76
3 0.32 0.11 -0.44
4 0.18 0.54  1.17
5 0.95 0.89 -1.60
6 0.95 0.60 -0.48

edited Mar 27, 2014 at 21:29

answered Mar 27, 2014 at 21:03

Rich Scriven

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1 Comment

purpleblau Over a year ago

Hey man, thank you very much for your effort. Although your answer doesn't really work on my example. But maybe it has to do with my dataframe. Josh's approach worked perfectly. I don't really know why his worked. But I learned a lot from yours too.

sunny singh · Accepted Answer · 2014-03-27 21:04:32Z

0

#include<iostream>
#include<conio.h>
#include<math.h>

using namespace std;

int main()
{
    float a=1.00634;
    float p=round(a*100); float q=p/100;
    printf("%.2f",q);   
    getch();
    return 0;
    }

answered Mar 27, 2014 at 21:04

sunny singh

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1 Comment

purpleblau Over a year ago

appreciate your answer, sorry, maybe my bad. My question is R related. not C++.

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