if-statement - 是否有 R 函数使用 If else 或

Question

我正在尝试根据以下条件在我的数据中创建一个变量：

x y Z S T G
1 0 1 0 1 0
1 0 0 0 0 0 
1 1 1 0 0 0 
1 1 1 1 1 1

if x=1 then 1, 
if y=1 then 2 if s=1 then 3, 
if t=1 then 4 if G=1 then 5 if X==y==z==1 then 6 and so on.

请告诉我如何使用 if else 来写这个

标签： if-statement

Answer 1

使用if else？您可以在没有 if else 的情况下计算它：

v <- 1:6
# this vector should give each column a the value
# 1 2 3 ... 6

# the most tedious part is to get your notes into a the R terminal
# as an R matrix.

# I used the fact that the string in R can span multiple lines:

s <- "x y Z S T G
1 0 1 0 1 0
1 0 0 0 0 0 
1 1 1 0 0 0 
1 1 1 1 1 1"
# it looks like this:

s
## [1] "x y Z S T G\n1 0 1 0 1 0\n1 0 0 0 0 0 \n1 1 1 0 0 0 \n1 1 1 1 1 1"

# after trying long around with the base R functions
# which led to errors and diverse problems, I found the most elegant way
# to transform this string into a matrix-like tabular form
# is to use tidyverse's read_delim().
# install.packages("tidyverse")

# load tidyverse:
require(tidyverse) # or: library(tidyverse)

tb <- read_delim(s, delim=" ") ## it complains about parsing failues, but
tb
# A tibble: 4 x 6
      x     y     Z     S     T     G
  <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1     1     0     1     0     1     0
2     1     0     0     0     0     0
3     1     1     1     0     0     0
4     1     1     1     1     1     1
# so it is read correctly in!



# what you want to do actually is
# to multiply each row with `v` and sum this result:

tb[1, ]
# A tibble: 1 x 6
      x     y     Z     S     T     G
  <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1     1     0     1     0     1     0

# you do:

v * tb[1, ]

  x y Z S T G
1 1 0 3 0 5 0

# if you build sum with this, then you get your desired numbers

sum(v * tb[1, ])
## [1] 9


# row-wise manipulation of matrix/data.frame/tibbles you do by 
apply(tb, MARGIN=1, FUN=function(row) v * row)
  [,1] [,2] [,3] [,4]
x    1    1    1    1
y    0    0    2    2
Z    3    0    3    3
S    0    0    0    4
T    5    0    0    5
G    0    0    0    6

# very often such functions flip the results, so flip it back 
# by the transpose function `t()`:
t(apply(tb, MARGIN=1, FUN=function(row) v * row))
     x y Z S T G
[1,] 1 0 3 0 5 0
[2,] 1 0 0 0 0 0
[3,] 1 2 3 0 0 0
[4,] 1 2 3 4 5 6

# to get directly the sum by row, do:
apply(tb, MARGIN=1, FUN=function(row) sum(v * row))
## [1]  9  1  6 21

# these are the values you wanted, isn't it?

# I see now, that 

tb * v    # by using vectorization of R

     x y Z S T G
[1,] 1 0 3 0 5 0
[2,] 1 0 0 0 0 0
[3,] 1 2 3 0 0 0
[4,] 1 2 3 4 5 6

# therfore the rowSums are:

rowSums(tb * v)
## [1]  9  1  6 21

因此，这是人们经常获得解决方案的通常（混乱）方式。

最后，归结为这一点（通常你会在 Stack Overflow 中找到类似的简短答案）：

简短的回答

require(tidyverse)

s <- "x y Z S T G
1 0 1 0 1 0
1 0 0 0 0 0 
1 1 1 0 0 0 
1 1 1 1 1 1"

tb <- read_delim(s, delim=" ")
rowSums(tb * v)

这就是 R 的美妙之处：如果您确切知道该做什么，只需 1-3 行代码（或更多）......