如何创建一个从数据框中提取列并使用R创建另一个函数的函数？

Question

I need to create a function to make my workflow better. I already create a code, but I need to repeat several times and change the number according to the mat[i] columns. I mean, if I'm going to create another matrix for mat2 , I need to replace in all my code, the number 1 to 2 , for mat3 : 2 to 3 , for mat4 : 4 to 5 .

I have the data frame below (it is not my real data):

year <- c(rep(1998:2001, 4))
Age <- c(rep(15:18, 4))
mat1 <- c(rep(0.01, 16))
mat2 <- c(rep(0.012, 16))

df <- data.frame(year, Age, mat1, mat2)

I need to create n final numeric matrices, one for each mat(i) , (i = 1, 2,...,n). The rows would be the age and in the columns year .

I got the result that I wanted using the code below:

library(dplyr)
mat1 <- #selecting just intensities of order 1 and creating matrices
  dplyr::select(df, Age, year, mat1) #choosig the variables that I want to keep %>% 
  spread(year, mat1)

names(mat1)[c(2:5)] <- paste0("year ", names(mat1[2:5])) #change colnames as it is in the model
mat1[ ,1] <- paste0("age ", mat1[,1]) #alter the row from column "age"

mat_oe1 <- data.matrix(mat1[2:5])
dimnames(mat_oe1) <- list(c(mat1[,1]), #row names
                        c(names(mat1[2:5])))#columns names

#Saving as txt to read in the model later
write.table(mat_oe1, file = "mat_oe1.txt", sep = "\t",
            row.names = T, col.names = T)

The final result (which must be equal this to be inserted in a model that is already created) that I got is for mat1 . For example, this result below it is for all the results for the variable mat1 :

view(mat1)
        year 1998  year 1999  year 2000  year 2001
age 15   0.01        0.01       0.01       0.01
age 16   0.01        0.01       0.01       0.01
age 17   0.01        0.01       0.01       0.01
age 18   0.01        0.01       0.01       0.01


view(mat2)
        year 1998  year 1999    year 2000    year 2001
age 15   0.012        0.012       0.012       0.012
age 16   0.012        0.012       0.012       0.012
age 17   0.012        0.012       0.012       0.012
age 18   0.012        0.012       0.012       0.012

Answer 1

Is this what you need

library(reshape2)
df$mat <- df$mat1+df$mat2
dcast(df,Age~year,value.var = "mat",sum)

EDIT

#df that you provided
structure(list(year = c(1998L, 1999L, 2000L, 2001L, 1998L, 1999L, 
2000L, 2001L, 1998L, 1999L, 2000L, 2001L, 1998L, 1999L, 2000L, 
2001L), Age = c(15L, 16L, 17L, 18L, 15L, 16L, 17L, 18L, 15L,   
16L, 17L, 18L, 15L, 16L, 17L, 18L), mat1 = c(0.1, 0.1, 0.1, 0.1, 
0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1), 
mat2 = c(0.12, 0.12, 0.12, 0.12, 0.12, 0.12, 0.12, 0.12, 
0.12, 0.12, 0.12, 0.12, 0.12, 0.12, 0.12, 0.12)), .Names = c("year", 
"Age", "mat1", "mat2"), row.names = c(NA, -16L), class = "data.frame")    

first seperate the df into two dataframe by column which contains mat and no mat

library(reshape2)
library(zoo)

df_no_mat <-  df[ , -which(names(df) %in% grep("mat", names(df), value = TRUE))]
#df_no_mat OUTPUT head
year    Age
1998    15
1999    16
2000    17
2001    18
1998    15
1999    16

df_mat <-  df[ , which(names(df) %in% grep("mat", names(df), value = TRUE))]
#df_mat OUTPUT head
mat1    mat2
 0.1    0.12
 0.1    0.12
 0.1    0.12
 0.1    0.12
 0.1    0.12
 0.1    0.12

Loop and spread which saves the df in a list

datalist = list()

for(i in names(df_mat)){
    df_no_mat$i <- df_mat[,i]
    datalist[[i]] <- dcast(df_no_mat,Age~paste0('year ',year),value.var = "i",mean)
    datalist[[i]] <- na.aggregate(datalist[[i]]) 
}

Access you df created for mats by using datalist$mat1 or datalist[[1]] and datalist$mat2 or datalist[[2]]

#Output
$mat1
Age year 1998   year 1999   year 2000   year 2001
15  0.01    0.01    0.01    0.01
16  0.01    0.01    0.01    0.01
17  0.01    0.01    0.01    0.01
18  0.01    0.01    0.01    0.01

$mat2
Age year 1998   year 1999   year 2000   year 2001
15  0.012   0.012   0.012   0.012
16  0.012   0.012   0.012   0.012
17  0.012   0.012   0.012   0.012
18  0.012   0.012   0.012   0.012