Skip to content
Back to formatted view

Raw Message

Message-ID: <CAPvBnPFY=xmCkWKbdvBd7tjqyDjmYCGdtr8pZod8D-dcCU=tsg@mail.gmail.com>
Date: 2011-10-19T23:44:03Z
From: Henrique Dallazuanna
Subject: replacing percentage of values in data frame
In-Reply-To: <1319067499329-3920484.post@n4.nabble.com>

Try this:

data$V6[sample(nrow(data), ceiling(length(data$V6) * 0.2))] <- 1

On Wed, Oct 19, 2011 at 9:38 PM, a217 <ajn21 at case.edu> wrote:
> I've been looking for how to change a certain percentage of values in a data
> frame, but I've been struggling to find information in R.
>
> For example:
>
> #################example data##############
>> data
> ? ? ?V1 ? ?V2 ? ?V3 ?V4 V5 ?V6 V7
> 1 ? chr1 ? 500 ? 500 CHH ?0 0.5 ?+
> 2 ? chr1 ? 550 ? 550 CHH ?0 0.0 ?+
> 3 ? chr2 ? 700 ? 700 CHH ?0 0.0 ?+
> 4 ? chr2 ?1000 ?1000 CHH ?0 0.0 ?+
> 5 ? chr3 ? 100 ? 100 CHH ?0 0.0 ?+
> 6 ? chr4 ? 450 ? 450 ?CG ?0 0.0 ?+
> 7 ? chr5 ? 450 ? 450 CHH ?0 0.0 ?+
> 8 ? chr5 50034 50034 CHG ?0 0.0 ?+
> 9 ? chr7 50055 50055 CHG ?0 0.0 ?+
> 10 chr10 50063 50063 CHH ?0 0.0 ?+
>
>> dput(data)
> structure(list(V1 = structure(c(1L, 1L, 3L, 3L, 4L, 5L, 6L, 6L,
> 7L, 2L), .Label = c("chr1", "chr10", "chr2", "chr3", "chr4",
> "chr5", "chr7"), class = "factor"), V2 = c(500L, 550L, 700L,
> 1000L, 100L, 450L, 450L, 50034L, 50055L, 50063L), V3 = c(500L,
> 550L, 700L, 1000L, 100L, 450L, 450L, 50034L, 50055L, 50063L),
> ? ?V4 = structure(c(3L, 3L, 3L, 3L, 3L, 1L, 3L, 2L, 2L, 3L), .Label =
> c("CG",
> ? ?"CHG", "CHH"), class = "factor"), V5 = c(0L, 0L, 0L, 0L,
> ? ?0L, 0L, 0L, 0L, 0L, 0L), V6 = c(0.5, 0, 0, 0, 0, 0, 0, 0,
> ? ?0, 0), V7 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
> ? ?1L), .Label = "+", class = "factor")), .Names = c("V1", "V2",
> "V3", "V4", "V5", "V6", "V7"), class = "data.frame", row.names = c(NA,
> -10L))
>>
> ############################
>
> Say for instance, I'd like to change 20% of values in column 6 to '1'
> instead of zero or whatever value may be currently present. How would I
> approach this?
>
> I am working with a large data frame and I need to replace values in one of
> the columns for 10-20% of the entire dataset. I hope what I am trying to
> convey is understandable to you.
>
> --
> View this message in context: http://r.789695.n4.nabble.com/replacing-percentage-of-values-in-data-frame-tp3920484p3920484.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Henrique Dallazuanna
Curitiba-Paran?-Brasil
25? 25' 40" S 49? 16' 22" O