CODEWITHSUNDEEP
rmatrixdataframe
J Choi
J Choi

Reputation: 67

How to create the matrix from data frame?

i have a data frame as follows, I want to create a matrix, where the average sleep duration hour (SLP) is shown according to 3 recruiting site(site) and 5 recruiting year(year).

SLP site year 
8.6  1   2008  
7.2  1   2005  
6.4  2   2006  
9.5  3   2007  
6.1  2   2009  
3.6  1   2005 
8.6  1   2008  
7.2  1   2005  
6.4  2   2006  
9.5  3   2007  
6.1  2   2009  
5.1  3   2008 
2.1  2   2006

My desired output is:

       1      2      3 
2005  6.00    -      -
2006   -     4.97    -
2007   -      -     9.5
2008  8.60    -     5.1
2009   -     6.10    -

Column name is variable of site, row name is variable of year and values in each cell is average of SLP. How do I do this?

Upvotes: 3

Views: 9338

Answers (4)

G. Grothendieck
G. Grothendieck

Reputation: 269481

Here are a few different solutions which use no packages:

1) tapply This uses no packages. It produces a "matrix" output with NA values for the empty cells:

tapply(DF$SLP, DF[c("year", "site")], mean)

giving:

      site
year     1        2   3
  2005 6.0       NA  NA
  2006  NA 4.966667  NA
  2007  NA       NA 9.5
  2008 8.6       NA 5.1
  2009  NA 6.100000  NA

2) aggregate/xtabs This uses aggregate + xtabs. This creates an object of class c("xtabs", "table") with zero values for empty cells:

fo <- SLP ~ year + site
xtabs(fo, aggregate(fo, DF, mean))

giving;

      site
year          1        2        3
  2005 6.000000 0.000000 0.000000
  2006 0.000000 4.966667 0.000000
  2007 0.000000 0.000000 9.500000
  2008 8.600000 0.000000 5.100000
  2009 0.000000 6.100000 0.000000

3) aggregate/reshape This also uses aggregate but uses reshape rather than xtabs. It gives a data frame r with NA's for empty cells. The last line makes the column names consistent with the prior solutions and could be omitted if this were not important.

 ag <- aggregate(SLP ~ site + year, DF, mean)
 r <- reshape(ag, dir = "wide", idvar = "year", timevar = "site")
 names(r) <- sub(".*[.]", "", names(r))

giving:

> r
  year   1        2   3
1 2005 6.0       NA  NA
3 2006  NA 4.966667  NA
5 2007  NA       NA 9.5
2 2008 8.6       NA 5.1
4 2009  NA 6.100000  NA

Note: The input DF in reproducible form used is:

DF <- structure(list(SLP = c(8.6, 7.2, 6.4, 9.5, 6.1, 3.6, 8.6, 7.2, 
6.4, 9.5, 6.1, 5.1, 2.1), site = c(1L, 1L, 2L, 3L, 2L, 1L, 1L, 
1L, 2L, 3L, 2L, 3L, 2L), year = c(2008L, 2005L, 2006L, 2007L, 
2009L, 2005L, 2008L, 2005L, 2006L, 2007L, 2009L, 2008L, 2006L
)), .Names = c("SLP", "site", "year"), class = "data.frame", row.names = c(NA, 
-13L))

Upvotes: 4

Adam
Adam

Reputation: 444

another solution

library(tidyr)
library(dplyr)

df%>% 
  group_by(year, site) %>%
    summarise(m=mean(SLP)) %>%
  spread(site, m )%>%
as.matrix()

Upvotes: 2

lmo
lmo

Reputation: 38500

Building on @g-grothendieck's use of xtabs, we can combine this with table and ifelse to return the same result.

# get a count of the number of observations per matrix cell (filling 0s with 1)
tempTab <- ifelse(with(df, table(year, + site)) == 0, 1, with(df, table(year, + site)))

tempTab

year   1 2 3
  2005 3 1 1
  2006 1 3 1
  2007 1 1 2
  2008 2 1 1
  2009 1 2 1

Now use xtabs, which returns a sum of the values when multiple observations are in a cell and divide by tempTab to get the mean.

xtabs(SLP ~ year + site, df) / tempTab
      site
year          1        2        3
  2005 6.000000 0.000000 0.000000
  2006 0.000000 4.966667 0.000000
  2007 0.000000 0.000000 9.500000
  2008 8.600000 0.000000 5.100000
  2009 0.000000 6.100000 0.000000

Upvotes: 0

akrun
akrun

Reputation: 887038

We can use acast

library(reshape2)
acast(df1, year~site, value.var="SLP", mean)

Or using tapply from base R

with(df1, tapply(SLP, list(year, site), FUN = mean))

Upvotes: 4

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