CODEWITHSUNDEEP
r
rookiestudent069
rookiestudent069

Reputation: 1

derive additional column based on existing column

                       year
  ID                   2005  2006  2007  2008  2009  2010  2011  2012
  Altstadt/Nord        17821 17947 17764 17301 17211 17337 17421 17330
  Altstadt/Süd         26706 26778 26693 26218 26657 26443 26677 26722
  Bayenthal             8117  8388  8411  8455  8479  8834  8981  8911
  Bickendorf           16350 16303 16383 16321 16351 16429 16692 16813

I have created this table using: xtabs(Einwohneranzahl~raum+jahr,subsetHH).

Now i want it as a subset in order to derive an addition column (average on all years).

Upvotes: 0

Views: 42

Answers (3)

Chris Ruehlemann
Chris Ruehlemann

Reputation: 21400

If speed of execution is not an issue (rowSums, rowMeans are faster), you can also use apply (using @GKi's data) as well as sum(if you want to add up the values):

x$new <- apply(x[-1], 1, sum)

Result:

x
             ID X2005 X2006 X2007 X2008 X2009 X2010 X2011 X2012    new
1 Altstadt/Nord 17821 17947 17764 17301 17211 17337 17421 17330 140132
2  Altstadt/Süd 26706 26778 26693 26218 26657 26443 26677 26722 212894
3     Bayenthal  8117  8388  8411  8455  8479  8834  8981  8911  68576
4    Bickendorf 16350 16303 16383 16321 16351 16429 16692 16813 131642

If it is the average you want to compute replace sum by mean.

Upvotes: 0

GKi
GKi

Reputation: 39647

You can use rowMeans to add a mean over all years like:

x$Mean <- rowMeans(x[-1])
x
#             ID X2005 X2006 X2007 X2008 X2009 X2010 X2011 X2012     Mean
#1 Altstadt/Nord 17821 17947 17764 17301 17211 17337 17421 17330 17516.50
#2  Altstadt/Süd 26706 26778 26693 26218 26657 26443 26677 26722 26611.75
#3     Bayenthal  8117  8388  8411  8455  8479  8834  8981  8911  8572.00
#4    Bickendorf 16350 16303 16383 16321 16351 16429 16692 16813 16455.25

Data:

x <- read.table(header=TRUE, text=" ID                   2005  2006  2007  2008  2009  2010  2011  2012
  Altstadt/Nord        17821 17947 17764 17301 17211 17337 17421 17330
  Altstadt/Süd         26706 26778 26693 26218 26657 26443 26677 26722
  Bayenthal             8117  8388  8411  8455  8479  8834  8981  8911
  Bickendorf           16350 16303 16383 16321 16351 16429 16692 16813")

Upvotes: 1

Charlie
Charlie

Reputation: 198

The rowSums() function can do this. It's a bit easier if you turn it into a data frame first.

library(dplyr)
# Make Cross tab and turn into data frame
cross_tab <- as.data.frame(xtabs(Einwohneranzahl~raum+jahr,subsetHH))

# Add a column that is the sum of each row
mutate(cross_tab, all_years = rowSums(cross_tab))

Upvotes: 0

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