Combine some columns of two matrices but with common information transposed - r

Question

I have the following two matrices:
matrix1 (first 10 rows and only some relevant columns):
Prod_Y2010 Prod_Y2011 Prod_Y2012 Prod_Y2013 Prod_Y2014 Place
1 6101 5733 5655 5803 5155 3
2 4614 4513 4322 5211 4397 1
3 5370 5295 4951 5145 4491 3
4 5689 5855 5600 5787 4848 1
5 3598 3491 3462 3765 3094 2
6 6367 6244 5838 6404 5466 7
7 2720 2635 2465 2917 2623 2
8 5077 5113 4456 5503 4749 8
9 5260 5055 4512 5691 4876 2
10 4771 4583 4202 5266 4422 2
where each column is grassland productivity from years 2010 to 2014, and the last column is the place where productivity was measured.
and matrix2:
Year Rain_Place1 Rain_Place2 Rain_Place3 Rain_Place7 Rain_Place8
11 2010 123.0 361.0 60.5 469.7 492.3
12 2011 45.5 404.4 224.8 395.4 417.3
13 2012 318.7 369.4 115.7 322.6 385.8
14 2013 93.2 378.4 155.5 398.2 413.1
15 2014 216.8 330.0 31.0 344.0 387.5
where for each of the same 5 years of matrix1 (which are the rows in matrix 2) I have data on the rainfall for each place.
I do not see how to proceed in R to join the information of the two matrices in such a way that my matrix1 has a series of additional columns intercalated (or interspersed) with the corresponding rain values matching the corresponding years and places. That is, what I need is a new matrix1 such as:
Prod_Y2010 Rain_Y2010 Prod_Y2011 Rain_Y2011 Prod_Y2012 Rain_Y2012 ... Place
1 6101 60.5 5733 224.8 5655 115.7 3
2 4614 123.0 4513 45.5 4322 318.7 1
3 5370 60.5 5295 224.8 4951 115.7 3
4 5689 123.0 5855 45.5 5600 318.7 1
5 3598 361.0 3491 404.4 3462 369.4 2
... ... ... ... ... ... ... ...
Of course the order is not important to me: if all the Rainfall columns are added as new columns at the right end of matrix1, that would be fine anyway.
Needless to say, my real matrices are several thousands rows long, and the number of years is 15.

Answer 1

I would second #jazzurro's comment- reformatting your data to long format would likely make it easier to work with for analysis etc. However, if you want to keep it using the wide format here is a way that might work- it uses the reshape2 and plyr libraries.
Given these data frames (dput() output of your data frames above, only included for reproducibility):
m1<-structure(list(Prod_Y2010 = c(6101L, 4614L, 5370L, 5689L, 3598L,
6367L, 2720L, 5077L, 5260L, 4771L), Prod_Y2011 = c(5733L, 4513L,
5295L, 5855L, 3491L, 6244L, 2635L, 5113L, 5055L, 4583L), Prod_Y2012 = c(5655L,
4322L, 4951L, 5600L, 3462L, 5838L, 2465L, 4456L, 4512L, 4202L
), Prod_Y2013 = c(5803L, 5211L, 5145L, 5787L, 3765L, 6404L, 2917L,
5503L, 5691L, 5266L), Prod_Y2014 = c(5155L, 4397L, 4491L, 4848L,
3094L, 5466L, 2623L, 4749L, 4876L, 4422L), Place = c(3L, 1L,
3L, 1L, 2L, 7L, 2L, 8L, 2L, 2L)), .Names = c("Prod_Y2010", "Prod_Y2011",
"Prod_Y2012", "Prod_Y2013", "Prod_Y2014", "Place"), class = "data.frame", row.names = c(NA,
-10L))
m2<-structure(list(Year = 2010:2014, Rain_Place1 = c(123, 45.5, 318.7,
93.2, 216.8), Rain_Place2 = c(361, 404.4, 369.4, 378.4, 330),
Rain_Place3 = c(60.5, 224.8, 115.7, 155.5, 31), Rain_Place7 = c(469.7,
395.4, 322.6, 398.2, 344), Rain_Place8 = c(492.3, 417.3,
385.8, 413.1, 387.5)), .Names = c("Year", "Rain_Place1",
"Rain_Place2", "Rain_Place3", "Rain_Place7", "Rain_Place8"), class = "data.frame", row.names = c("11",
"12", "13", "14", "15"))
To get the place number from the column names in your rain data frame to use in a later join:
rename <- function(x) {
y <- substr(x, nchar(x), nchar(x))
return(y)
}
Edit: Here is a better rename function, that should work with more than 9 places (modified from an answer here):
rename <- function(x) {
y <- unlist(regmatches(x, gregexpr('\\(?[0-9,.]+', x)))
return(y)
}
sapply(names(m2[2:ncol(m2)]), FUN = rename)
names(m2) <- c(names(m2)[1], sapply(names(m2[2:ncol(m2)]), FUN = rename))
> m2
Year 1 2 3 7 8
1 2010 123.0 361.0 60.5 469.7 492.3
2 2011 45.5 404.4 224.8 395.4 417.3
3 2012 318.7 369.4 115.7 322.6 385.8
4 2013 93.2 378.4 155.5 398.2 413.1
5 2014 216.8 330.0 31.0 344.0 387.5
Melt the rain data frame:
m3<-melt(m2, id.vars = "Year", variable.name = "Place", value.name = "Rain")
> head(m3)
Year Place Rain
1 2010 1 123.0
2 2011 1 45.5
3 2012 1 318.7
4 2013 1 93.2
5 2014 1 216.8
6 2010 2 361.0
Reshape the melted data frame to allow for a join by "Place", and treat "Place" as a character rather than a factor:
m4<-reshape(m3, idvar = "Place", timevar = "Year", direction = "wide")
m4$Place <- as.character(m4$Place)
> m4
Place Rain.2010 Rain.2011 Rain.2012 Rain.2013 Rain.2014
1 1 123.0 45.5 318.7 93.2 216.8
6 2 361.0 404.4 369.4 378.4 330.0
11 3 60.5 224.8 115.7 155.5 31.0
16 7 469.7 395.4 322.6 398.2 344.0
21 8 492.3 417.3 385.8 413.1 387.5
Finally, join this melted/reshaped data frame to your "Prod" data frame.
m5<-join(m1, m4, by = "Place")
> m5
Prod_Y2010 Prod_Y2011 Prod_Y2012 Prod_Y2013 Prod_Y2014 Place Rain.2010 Rain.2011 Rain.2012 Rain.2013 Rain.2014
1 6101 5733 5655 5803 5155 3 60.5 224.8 115.7 155.5 31.0
2 4614 4513 4322 5211 4397 1 123.0 45.5 318.7 93.2 216.8
3 5370 5295 4951 5145 4491 3 60.5 224.8 115.7 155.5 31.0
4 5689 5855 5600 5787 4848 1 123.0 45.5 318.7 93.2 216.8
5 3598 3491 3462 3765 3094 2 361.0 404.4 369.4 378.4 330.0
6 6367 6244 5838 6404 5466 7 469.7 395.4 322.6 398.2 344.0
7 2720 2635 2465 2917 2623 2 361.0 404.4 369.4 378.4 330.0
8 5077 5113 4456 5503 4749 8 492.3 417.3 385.8 413.1 387.5
9 5260 5055 4512 5691 4876 2 361.0 404.4 369.4 378.4 330.0
10 4771 4583 4202 5266 4422 2 361.0 404.4 369.4 378.4 330.0