Hi all,
I think this should be an easy question for the guru's out here.
I have this large data frame (2.500.000 rows, 15 columns) and I want to add
a column named "SEGMENT" to it.
The first 5% rows (first 125.000 rows) should have the value "Top 5%" in the
SEGMENT column
Then the rows from 5% to
Hello,
Combine quantile() with findInterval(). Something like the following.
# sample data
x <- rnorm(100)
val <- c("Bottom 50", "20 to 50", "5 to 20", "Top 5%")
qq <- quantile(x, probs = c(0, 0.50, 0.70, 0.95, 1))
idx <- findInterval(x, qq)
val[idx]
Hope this helps,
Rui Barradas
Em 31-07
MENT
#495 1 0.3571848 Bottom 50
#496 9 -1.1971854 Bottom 50
#497 5 0.3544896 Bottom 50
#498 8 -0.1562356 Bottom 50
#499 8 -0.2994321 Bottom 50
#500 8 -0.4170319 Bottom 50
A.K.
- Original Message -
From: Dark
To: r-help@r-project.org
Cc:
Sent: Wednesday, July 31, 2013 5:37 AM
Subjec
119 119 2.589574 20 to 50
12 12 2.512276 Top 5%
Do you have a solution for this?
- Original Message -
From: arun
To: Dark
Cc: R help
Sent: Wednesday, July 31, 2013 7:48 AM
Subject: Re: [R] Add a column to a data frame with value based on the
percentile of the row
Hi,
May
Hello,
Sorry, that should be 0.80, not 0.70.
qq <- quantile(x, probs = c(0, 0.50, 0.80, 0.95, 1))
Rui Barradas
Em 31-07-2013 12:22, Rui Barradas escreveu:
Hello,
Combine quantile() with findInterval(). Something like the following.
# sample data
x <- rnorm(100)
val <- c("Bottom 50", "20
Works like a charm, thanks a lot!
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Hi Arun Kirshna,
I have tested your method and it will work for me.
I only run into one problem. Before I want to do this operation I have
sorted my data frame so my rownumbers ar not subsequent.
You can see if you first order your example data frame like:
dat1 <- dat1[order(-dat1$value),]
head(
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