在 R 中使向量长度相同

Question

This task is not as simple as the title suggests. It's better for me to just use a lame example to explain what I want. I have two vectors x and y...

x <- c("Description 1 2 3 4","5 6 7 8 9 10 11 12","13 14 15 16 17","18 19 20 21 22","23 24 25","26 27 28","","29","30 31","Tot") 
y <- c("Minutes","","","35","60 60 30","60 60","","15","60 60","440")
rbind(x,y)

x "Description 1 2 3 4"     "5 6 7 8 9 10 11 12" "13 14 15 16 17" "18 19 20 21 22" "23 24 25" "26 27 28" ""   "29" "30 31" "Tot"
y "Minutes"                 ""                   ""               "35"             "60 60 30" "60 60"    ""   "15" "60 60" "440"

I need the minutes to align for the specific days from a pdf table. It helps to see what I want if you look at x and y combined together (above). For each "column", if the minutes have more than one day above them, I need to shift the minutes to the last day. For example, on days 18-22, we see there were 35 minutes for one of those days... I need to shift those 35 minutes to correspond to the 22nd day of the month. For each day that doesn't have any minutes I need to give that day a value of NA. The result should look like the following...

result <- data.frame(rbind(seq(1:31),c(rep(NA,21),35,60,60,30,NA,60,60,15,60,60)))
result

X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 X15 X16 X17 X18 X19 X20 X21 X22 X23 X24 X25 X26 X27 X28 X29 X30 X31
1  1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  30  31
2 NA NA NA NA NA NA NA NA NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  35  60  60  30  NA  60  60  15  60  60

Any help appreciated!

Update:

I was able to the solve the problem with the following code...

z <- rbind(x,y)
z <- z[,-ncol(z)]
result <- lapply(1:ncol(z),function(x){
    print(x)
    col <- z[,x]
    row1 <- do.call("c",str_extract_all(col[1],"\\(?[0-9,.]+\\)?"))
    row2 <- do.call("c",str_extract_all(col[2],"\\(?[0-9,.]+\\)?"))
    if(length(row2) == 0) {
        w <- rbind(row1,rep(NA,length(row1)))
    } else {
        w <- rbind(row1,c(rep(NA,length(row1)-length(row2)),row2))
    }
    w
})
do.call("cbind",result)

Would be interested in seeing other people'e solutions still.

Answer 1

Here is another approach - see what you think.

# First four lines are from your original post
x <- c("Description 1 2 3 4","5 6 7 8 9 10 11 12","13 14 15 16 17","18 19 20 21 22","23 24 25","26 27 28","","29","30 31","Tot")
y <- c("Minutes","","","35","60 60 30","60 60","","15","60 60","440")
z <- rbind(x,y)
z <- z[,-ncol(z)]

# Function to extract numbers
match_fxn <- function(x) { 
  matches <- regmatches(x, gregexpr("[[:digit:]]+", x))
  as.numeric(unlist(matches))
}

# Create nested matrix of extracted numbers, 
mat <- apply(z, 1:2, match_fxn)

# Force lengths to be same between rows 1 and 2 (adds NAs)
min_list <- mapply(`length<-`, x = mat[2,], Var = lengths(mat[1,]))

# Sort so that NAs are in the beginning
min_sort <- lapply(min_list, function(x) c(x[is.na(x)], x[!is.na(x)]))

# Add rows 1 and 2 to show final result
rbind(unlist(mat[1,]), unlist(min_sort))

Output

     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16] [,17] [,18] [,19] [,20] [,21] [,22] [,23] [,24] [,25] [,26] [,27] [,28] [,29] [,30] [,31]
[1,]    1    2    3    4    5    6    7    8    9    10    11    12    13    14    15    16    17    18    19    20    21    22    23    24    25    26    27    28    29    30    31
[2,]   NA   NA   NA   NA   NA   NA   NA   NA   NA    NA    NA    NA    NA    NA    NA    NA    NA    NA    NA    NA    NA    35    60    60    30    NA    60    60    15    60    60