Question

我有一个数据集，该数据集的变量ColumnStart标识第一列以计算平均值。我有第二个变量ColumnEnd，它标识该计算中的最后一列。对于第一行，我想计算第5列到第9列的平均值，第二行从第6列到第11列，等等。

输出为：

这是R中更新的dput：

structure(list(ID = c("AAA", "BBB", "CCC", "DDD"), ShortID = c("452L", 
"3L", "4L", "324L"), Name = c("PS1", "PS2", "PS3", "PS4"), Route = 
c("Internal", 
"External", "Internal", "Internal"), ColumnStart = c(7L, 7L, 
9L, 8L), ColumnEnd = c(9L, 11L, 13L, 10L), Date1 = c(1L, 5L, 
13L, 4L), Date2 = c(2L, 6L, 45L, 3L), Date3 = c(3L, 7L, 23L, 
2L), Date4 = c(4L, 8L, 65L, 1L), Date5 = c(5L, 8L, 34L, 3L), 
Date6 = c(6L, 9L, 23L, 5L), Date7 = c(7L, 6L, 54L, 6L), Date8 = c(7L, 
6L, 1L, 7L), Date9 = c(8L, 9L, 3L, 8L)), .Names = c("ID", 
"ShortID", "Name", "Route", "ColumnStart", "ColumnEnd", "Date1", 
"Date2", "Date3", "Date4", "Date5", "Date6", "Date7", "Date8", 
"Date9"), row.names = c(NA, -4L), class = c("tbl_df", "tbl", 
"data.frame"), spec = structure(list(cols = structure(list(ID = 
structure(list(), class = c("collector_character", 
"collector")), ShortID = structure(list(), class = 
c("collector_character", 
"collector")), Name = structure(list(), class = c("collector_character", 
"collector")), Route = structure(list(), class = c("collector_character", 
"collector")), ColumnStart = structure(list(), class = 
c("collector_integer", 
"collector")), ColumnEnd = structure(list(), class = 
c("collector_integer", 
"collector")), Date1 = structure(list(), class = c("collector_integer", 
"collector")), Date2 = structure(list(), class = c("collector_integer", 
"collector")), Date3 = structure(list(), class = c("collector_integer", 
"collector")), Date4 = structure(list(), class = c("collector_integer", 
"collector")), Date5 = structure(list(), class = c("collector_integer", 
"collector")), Date6 = structure(list(), class = c("collector_integer", 
"collector")), Date7 = structure(list(), class = c("collector_integer", 
"collector")), Date8 = structure(list(), class = c("collector_integer", 
"collector")), Date9 = structure(list(), class = c("collector_integer", 
"collector"))), .Names = c("ID", "ShortID", "Name", "Route", 
"ColumnStart", "ColumnEnd", "Date1", "Date2", "Date3", "Date4", 
"Date5", "Date6", "Date7", "Date8", "Date9")), default = structure(list(), 
class = c("collector_guess", 
"collector"))), .Names = c("cols", "default"), class = "col_spec"))

Answer 1

这是一个基本的R解决方案，可以在计算平均值之前删除非数字列：

df$ave2 <- apply(df, 1, function(x) {
    y <- as.numeric(x[seq.int(x['ColumnStart'], x['ColumnEnd'])])
    mean(y[!is.na(y)])
    })

df
# A tibble: 4 x 16
  ID    ShortID Name  Route    ColumnStart ColumnEnd Date1 Date2 Date3 Date4 Date5 Date6 Date7 Date8 Date9 Average
  <chr> <chr>   <chr> <chr>          <int>     <int> <int> <int> <int> <int> <int> <int> <int> <int> <int>   <dbl>
1 AAA   452L    PS1   Internal           7         9     1     2     3     4     5     6     7     7     8     2  
2 BBB   3L      PS2   External           7        11     5     6     7     8     8     9     6     6     9     6.8
3 CCC   4L      PS3   Internal           9        13    13    45    23    65    34    23    54     1     3    39.8
4 DDD   324L    PS4   Internal           8        10     4     3     2     1     3     5     6     7     8     2

as.numeric尝试将值转换为numeric。如果不能，则返回NA。然后，我们删除NA的值并计算mean。

这里是一个单行版本，其工作原理相同，但在计算均值之前使用na.omit去除了NA值：

df$Average <- apply(df, 1, function(x) mean(na.omit(as.numeric(x[seq.int(x['ColumnStart'], x['ColumnEnd'])]))))

Answer 2

另一种方法，不一定建议

rowMeans(df*NA^!(col(df) >= df$ColumnStart & col(df) <= df$ColumnEnd), 
         na.rm = T)
# [1] 3.000000 7.142857 5.000000 3.333333 6.500000

说明：

col(df) >= df$ColumnStart & col(df) <= df$ColumnEnd是一个矩阵，在与TRUE，ColumnStart规范匹配的（i，j）索引处为ColumnEnd

NA^!(col(df) >= df$ColumnStart & col(df) <= df$ColumnEnd)是一个矩阵，该矩阵在其他地方的1和TRUE处为NA。用df对其进行互斥运算得到的矩阵与df相同，除了所有索引不符合ColumnStart和ColumnEnd规范的元素都是NA < / p>

现在我们可以使用其中的rowMeans和na.rm = T来获得所需的结果

根据列值计算平均值

2 个答案: