如何用我想要的数字替换findinterval函数的结果?下面是dput()输出:R中的findInterval函数(设置结果)
a=c(113,113,113,113,111,111,115,116,117,118,220,220)
b=c(113,113,113,113,111,111,115,116,117,118,220,220)
c=c(2,2,1,1,5,1,1,2,1,1,1,4)
d=c(2,2,12,12,15,12,12,2,12,12,12,14)
e=c(1,1,1,1,1,2,2,2,2,2,2,3)
f=c(20,30,25,35,45,55,60,65,70,75,75,80)
h=c("1A","1A","2A","3A","1A","5A","4A","4A","7A","7A","9A","9A")
i=c(12,16,17,19,20,15,18,17,17,13,14,15)
m=data.frame(a=a,b=b,c=c,d=d,e=e,f=f,h=h,i=i)
dput(m)
structure(list(a = c(113, 113, 113, 113, 111, 111, 115, 116,
117, 118, 220, 220), b = c(113, 113, 113, 113, 111, 111, 115,
116, 117, 118, 220, 220), c = c(2, 2, 1, 1, 5, 1, 1, 2, 1, 1,
1, 4), d = c(2, 2, 12, 12, 15, 12, 12, 2, 12, 12, 12, 14), e = c(1,
1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3), f = c(20, 30, 25, 35, 45, 55,
60, 65, 70, 75, 75, 80), h = structure(c(1L, 1L, 2L, 3L, 1L,
5L, 4L, 4L, 6L, 6L, 7L, 7L), .Label = c("1A", "2A", "3A", "4A",
"5A", "7A", "9A"), class = "factor"), i = c(12, 16, 17, 19, 20,
15, 18, 17, 17, 13, 14, 15)), .Names = c("a", "b", "c", "d",
"e", "f", "h", "i"), row.names = c(NA, -12L), class = "data.frame")
set.seed(5)
m$rand <- runif(nrow(m))
m[a==113,"i"] <- c(10,11,12)[1+findInterval(unlist(m[m$a==113,"rand",with=F]),c(0.25,0.50))]
有没有什么简单的方法,从具有所有这些信件在一个地方值向量绘图。例如,对于== 113 c(0.25,0.50),values = c(10,11,12)] [for a == 111 c(0.25,0.50,0.75),values = c(1,2, 3,4)] [对于== 115 c(0.25,0.50,0.75),values = c(1,2,3,4)]都在一个表或框架中,并且每当findinterval函数被使用时从这些绘制出来? i列应该被相关的值替换。我想要做的是从另一个文件读取值(例如c(10,11,12)),并在需要时放入findinterval函数。
如果您忽略大部分data.table语法和内存效率,为什么还要使用data.table?你想要的数字是什么? – mnel
它给了我错误,而不使用data.table。期望的数字是10和11(对于小于0.45的数字,得到数字10,对于大于0.45的rand,得到数字11)。我将为表格的其他部分使用相同的过程。 – Kaveh
你是从这些数字中拉出来的? – mnel