R 中的条件行总和

Question

    a   avalue  b   bvalue
1  12   yes     3   no
2  13   yes     3   yes
3  14   no      2   no
4  NA   no      1   no
5  16   NA      1   yes

I'm trying to count the total number of yes in each row, so the output would be like this:

   Count
1  12
2  16
3  0
4  0
5  1

Here is my solution which is not working. df$count <- rowSums(data[data(3) | data(5) == 'yes',c(2,4)], na.rm=TRUE)

Answer 1

Edit:

OP has edited the post to include headers on the input data, and judging from the comments, it seems that OP wants the solution to scale to multiple column pairs. Here's a solution in base R that should do that:

raw <- "
   a   avalue  b   bvalue
1  12   yes     3   no
2  13   yes     3   yes
3  14   no      2   no
4  NA   no      1   no
5  16   NA      1   yes "

df <- read.table(text = raw, header = TRUE)

use <- endsWith(colnames(df), "value")
df[use] <- ifelse(df[use] == "yes", TRUE, FALSE)
df[is.na(df)] <- 0
rowSums(df[use] * df[!use])
#>  1  2  3  4  5 
#> 12 16  0  0  1

^{Created on 2021-02-20 by the reprex package (v0.3.0)}

Original post:

Another take:

raw <- "1  12   yes     3   no
2  13   yes     3   yes
3  14   no      2   no
4  NA   no      1   no
5  16   NA      1   yes"

df <- read.table(text = raw)

suppressPackageStartupMessages({
  library(dplyr)
  library(tidyr)
})

df %>%
  setNames(c("row", "value_first", "use_first", "value_second", "use_second")) %>%
  pivot_longer(!row, names_to = c(".value", "column"), names_sep = "_") %>%
  replace_na(list(value = 0, use = "no")) %>%
  group_by(row) %>%
  summarise(total = sum(value * (use == "yes")))
#> # A tibble: 5 x 2
#>     row total
#> * <int> <dbl>
#> 1     1    12
#> 2     2    16
#> 3     3     0
#> 4     4     0
#> 5     5     1

^{Created on 2021-02-18 by the reprex package (v0.3.0)}

Answer 2

Or using base R, you can simply do element-wise multiplication for the rows that satisfy your condition on the value column, and then apply rowSums() :

raw <- "1  12   yes     3   no
2  13   yes     3   yes
3  14   no      2   no
4  NA   no      1   no
5  16   NA      1   yes"

df <- read.table(text = raw)

rowSums((!is.na(df[,c(3,5)])&df[,c(3,5)]=="yes") * df[,c(2,4)], na.rm=TRUE)
#> [1] 12 16  0  0  1

## Explanation:
# 1) Select relevant rows
(rows_select <- !is.na(df[,c(3,5)])&df[,c(3,5)]=="yes")
#>         V3    V5
#> [1,]  TRUE FALSE
#> [2,]  TRUE  TRUE
#> [3,] FALSE FALSE
#> [4,] FALSE FALSE
#> [5,] FALSE  TRUE

# 2) multiply by the columns with the data:
(rows_sel_val <- rows_select * df[,c(2,4)])
#>   V2 V4
#> 1 12  0
#> 2 13  3
#> 3  0  0
#> 4 NA  0
#> 5  0  1

# 3) Apply rowSums
rowSums(rows_sel_val, na.rm=TRUE)
#> [1] 12 16  0  0  1

^{Created on 2021-02-18 by the reprex package (v1.0.0)}

Answer 3

1) Create a new data frame df0 that has 0 where each NA in df is and then use the indicated formula on it. No packages are used.

df0 <- replace(df, is.na(df), 0)
transform(df, count = with(df0, a * (avalue == "yes") + b * (bvalue == "yes")))

giving:

   a avalue b bvalue count
1 12    yes 3     no    12
2 13    yes 3    yes    16
3 14     no 2     no     0
4 NA     no 1     no     0
5 16   <NA> 1    yes     1

2) or if there are more than just a and b then this gives the same result but handles any number of columns. ok picks out the a, b, etc. columns and,ok picks out the avalue, bvalue. etc. columns. Note that R will automatically recycle ok and !ok to a length equal to the number of columns.

ok <- c(TRUE, FALSE)
transform(df, count = rowSums(df[ok] * (df[!ok] == "yes"), na.rm = TRUE))

2a) Using the collapse package, a variation on (2) is to use num_vars and cat_vars which pick out the numeric and categorical columns.

Note that if any of the numeric columns are all NA then they must be set using NA_real_ or NA_integer_ and not just NA since num_vars is extracting columns by type. This can be checked by ensuring that logi_vars(df) has no columns (since an ordinary NA is logical) or else just use (2) if it is possible that any column is all NA.

library(collapse)

transform(df, count = rowSums(num_vars(df0) * (cat_vars(df0) == "yes"), na.rm = TRUE))

Note

The input in reproducible form is:

Lines <- "
    a   avalue  b   bvalue
1  12   yes     3   no
2  13   yes     3   yes
3  14   no      2   no
4  NA   no      1   no
5  16   NA      1   yes"
df <- read.table(text = Lines)

Answer 4

I'd tidy my data and calculate the sum by group, using the tidyverse:

library(tidyverse)
df<-read.table(text = "1  12   yes     3   no
2  13   yes     3   yes
3  14   no      2   no
4  NA   no      1   no
5  16   NA      1   yes")

bind_rows(df[1:3], setNames(df[c(1,4:5)], paste0("V",1:3))) %>%
group_by(V1, V3) %>%
summarise(sum(V2, na.rm = TRUE))

#> Groups:   V1 [5]
#>     V1 V3    `sum(V2, na.rm = TRUE)`
#>  <int> <chr>                   <int>
#>1     1 no                          3
#>2     1 yes                        12
#>3     2 yes                        16
#>4     3 no                         16
#>5     4 no                          1
#>6     5 yes                         1
#>7     5 <NA>                       16