control flow for calling other columns in function

问题

I'm trying to make a connection to other columns within a function given a conditional.

Essentially, I want to make a dataframe go from long to wide given a conditional, where those values in one column are NA relative to another column that has values in that same row, turn the NAs into a specific numeric.

Although the values assigned have to be column specific. So if 2010 has NAs whilst 2019 has a value, then return 16 otherwise, if 2019 has NAs when in that same row, 2010 has values, return 16.

what I have tried:

# A tibble: 26 x 4
   year  locality_id landcover  pland
   <chr> <chr>           <int>  <dbl>
 1 2010  L2228604           10 0.0645
 2 2010  L2228604           13 0.935 
 3 2010  L452817             8 0.0968
 4 2010  L452817             9 0.0323
 5 2010  L452817            12 0.613 
 6 2010  L452817            13 0.194 
 7 2010  L452817            14 0.0645
 8 2010  L596267             0 0.194 
 9 2010  L596267             9 0.0323
10 2010  L596267            11 0.0645
# ... with 16 more rows

p <- function(x){
    if(x == 'NA') {
        return(16)
    } else {
        17
    }
}

#wrong outcome
# A tibble: 26 x 4
   locality_id  pland `2010` `2019`
   <chr>        <dbl>  <dbl>  <dbl>
 1 L2228604    0.0645     17     NA
 2 L2228604    0.935      17     NA
 3 L452817     0.0968     17     NA
 4 L452817     0.0323     17     NA
 5 L452817     0.613      17     NA
 6 L452817     0.194      17     NA
 7 L452817     0.0645     17     NA
 8 L596267     0.194      17     NA
 9 L596267     0.0323     17     NA
10 L596267     0.0645     17     NA
# ... with 16 more rows

What I am expecting:

# A tibble: 26 x 4
   locality_id  pland `2010` `2019`
   <chr>        <dbl>  <int>  <int>
 1 L2228604    0.0645     10     17
 2 L2228604    0.935      13     17
 3 L452817     0.0968      8     17
 4 L452817     0.0323      9     17
 5 L9185766 0.54838710    16      8
 6 L9185766 0.19354839    16      9
 7 L9185766 0.03225806    16     13
 8 L9185766 0.16129032    16     14
 9 L9234578 1.00000000    16     12

reproducible code:

structure(list(year = c("2010", "2010", "2010", "2010", "2010", 
"2010", "2010", "2010", "2010", "2010", "2010", "2010", "2010", 
"2010", "2010", "2019", "2019", "2019", "2019", "2019", "2019", 
"2019", "2019", "2019", "2019", "2019"), locality_id = c("L2228604", 
"L2228604", "L452817", "L452817", "L452817", "L452817", "L452817", 
"L596267", "L596267", "L596267", "L596267", "L152650", "L910180", 
"L910180", "L910180", "L4791597", "L4791597", "L9149985", "L9149985", 
"L9149985", "L9185766", "L9185766", "L9185766", "L9185766", "L9185766", 
"L9234578"), landcover = c(10L, 13L, 8L, 9L, 12L, 13L, 14L, 0L, 
9L, 11L, 13L, 13L, 0L, 8L, 9L, 5L, 8L, 10L, 11L, 12L, 4L, 8L, 
9L, 13L, 14L, 12L), pland = c(0.0645161290322581, 0.935483870967742, 
0.0967741935483871, 0.032258064516129, 0.612903225806452, 0.193548387096774, 
0.0645161290322581, 0.193548387096774, 0.032258064516129, 0.0645161290322581, 
0.709677419354839, 1, 0.4375, 0.34375, 0.03125, 0.566666666666667, 
0.0333333333333333, 0.1, 0.0333333333333333, 0.866666666666667, 
0.0645161290322581, 0.548387096774194, 0.193548387096774, 0.032258064516129, 
0.161290322580645, 1)), row.names = c(NA, -26L), class = c("tbl_df", 
"tbl", "data.frame"))

回答1:

You can get the data in wide format and based on column name replace the NA value.

library(dplyr)
library(tidyr)

df %>%
  pivot_wider(names_from = year, values_from = landcover) %>%
  mutate(across(`2010`:`2019`, ~if(cur_column() == '2019') 
                               replace_na(.x, 17) else replace_na(.x, 16))) 


#   locality_id  pland `2010` `2019`
#   <chr>        <dbl>  <dbl>  <dbl>
# 1 L2228604    0.0645     10     17
# 2 L2228604    0.935      13     17
# 3 L452817     0.0968      8     17
# 4 L452817     0.0323      9     17
# 5 L452817     0.613      12     17
# 6 L452817     0.194      13     17
# 7 L452817     0.0645     14     17
# 8 L596267     0.194       0     17
# 9 L596267     0.0323      9     17
#10 L596267     0.0645     11     17
# … with 16 more rows

回答2:

There is a package called libr with a datastep() function that can do this easily. The datastep() function allows you to step through the data row by row, and make decisions based on the values of the columns. You can also create new columns on the fly.

Here is some code for your specific problem. I put your sample data in a tibble called dat:

library(libr)


final <- datastep(dat, 
                  keep = c("locality_id", "pland", "2010", "2019"), 
                  {
                    
                    if (year == "2010") {
                      "2010" <- landcover
                      "2019" <- 17
                    } else if (year == "2019") {
                      "2010" <- 16
                      "2019" <- landcover 
                    }

                  })

Above tibble final contains the following results:

# locality_id      pland 2010 2019
# 1     L2228604 0.06451613   10   17
# 2     L2228604 0.93548387   13   17
# 3      L452817 0.09677419    8   17
# 4      L452817 0.03225806    9   17
# 5      L452817 0.61290323   12   17
# 6      L452817 0.19354839   13   17
# 7      L452817 0.06451613   14   17
# 8      L596267 0.19354839    0   17
# 9      L596267 0.03225806    9   17
# 10     L596267 0.06451613   11   17
# 11     L596267 0.70967742   13   17
# 12     L152650 1.00000000   13   17
# 13     L910180 0.43750000    0   17
# 14     L910180 0.34375000    8   17
# 15     L910180 0.03125000    9   17
# 16    L4791597 0.56666667   16    5
# 17    L4791597 0.03333333   16    8
# 18    L9149985 0.10000000   16   10
# 19    L9149985 0.03333333   16   11
# 20    L9149985 0.86666667   16   12
# 21    L9185766 0.06451613   16    4
# 22    L9185766 0.54838710   16    8
# 23    L9185766 0.19354839   16    9
# 24    L9185766 0.03225806   16   13
# 25    L9185766 0.16129032   16   14
# 26    L9234578 1.00000000   16   12

来源：https://stackoverflow.com/questions/65865839/control-flow-for-calling-other-columns-in-function

标签

function

pivot

control-flow