我有这样的数据
df<-structure(list(data = structure(c(8L, 2L, 3L, 2L, 2L, 2L, 2L,
1L, 7L, 5L, 6L, 5L, 4L), .Label = c("1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0",
"2, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0",
"2, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0",
"2, 2, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0",
"2, 2, 2, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0",
"3, 2, 2, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0",
"M1yrtr", "Mitered"), class = "factor")), row.names = c(NA, -13L), class = "data.frame")
我想为每一行计算以下内容
例如,第二行是
2, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
我想计算一下
n =5
(-(2/n)*log2(2/n)) + (-(1/n)*log2(1/n)) +(-(1/n)*log2(1/n))+ (-(1/n)*log2(1/n))
对于第三个是
2, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
我会计算一下
(-(2/n)*log2(2/n)) + (-(2/n)*log2(2/n)) + (-(1/n)*log2(1/n))
所以输出看起来像这样
dfout<- structure(list(data = structure(c(8L, 2L, 3L, 2L, 2L, 2L, 2L,
1L, 7L, 5L, 6L, 5L, 4L), .Label = c("1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0",
"2, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0",
"2, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0",
"2, 2, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0",
"2, 2, 2, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0",
"3, 2, 2, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0",
"M1yrtr", "Mitered"), class = "factor"), X = structure(c(8L,
3L, 2L, 3L, 3L, 3L, 3L, 1L, 7L, 6L, 4L, 6L, 5L), .Label = c("0.2604594",
"1.03563", "1.168964", "2.020935", "2.077468", "2.204594", "M1yrtr",
"Mitered"), class = "factor")), class = "data.frame", row.names = c(NA,
-13L))
在R中,所有基本操作(加法减法,乘法,对数......)都是矢量化的。这意味着,例如,如果x
是一个向量,那么log(x)
只是成分log
函数,矿石1 / x
只是组成分裂。
因此,您可以执行以下操作:
x <- as.numeric(str_split(df[2, ], ", ", simplify = T))
n <- 5
sum((-(x[x > 0]/n)*log2(x[x > 0]/n)))
[1] 1.921928
如果要将其应用于所有行,可以使用sapply
函数,如下所示:
myfun <- function(x){
if (! grepl(",", x)) return(as.character(x))
n <- 5
y <- as.numeric(str_split(x, ", ", simplify = T))
as.character(sum((-(y[y > 0]/n)*log2(y[y > 0]/n))))
}
df$newcol <- sapply(df[,1], myfun)