【发布时间】:2015-05-18 19:55:13
【问题描述】:
Person <- c(1,2,3)
Age <- c(10,22,30)
Height <- c(140,185,160)
Weight <- c(65, 80, 75)
People <- data.frame(Person, Age, Height, Weight)
Age_cats_type1 [5-15], [20-30], [35-45]
Age_cats_type2 [8-13], [14-16], [18-40]
Height_cat_Type1 [100-120], [121-140], [141-186]
Height_cat_type2 [110-125], [126-145], [146-190]
Weight_cat_Type1 [50-60], [61-78], [79-85]
Weight_cat_Type2 [55-75], [76-90], [91-100]
对于 People[1,2](年龄=10),这适合 Age_cats_type1==1 和 Age_cats_type2==1。
对于 People[1,3] (height=140),这适合 Height_cat_Type1==2 和 Height_cat_Type2==2
现在我想为(Age_cats_type1==1)|(Age_cats_type1==2)、(Height_cats_type1==1)|(Height_cats_type1==2)、(Weight_cats_type1==1)|(Weight_cats_type1==2) 的间隔的每个唯一可能结果创建一个表格。
所需的输出应该类似于下面的黄色图像。 上表是对每个 invinterval 的可能性的总结
这与following question 密切相关,但是当您按照 BrodieG 概述的此处使用的代码进行操作时,第三次迭代会出现错误。
在这个例子中,我们在data.table中使用foverlaps
我使用了以下代码
library(intervals)
# create our limits
AGE_cats_type1 <- Intervals(
matrix(c(5, 15, 20, 30, 35, 40), ncol = 2, byrow = TRUE ),
closed = c( TRUE, T ),
type = "Z"
)
AGE_cats_type2 <- Intervals(
matrix(c(8, 13, 14, 16, 18, 40), ncol = 2, byrow = TRUE ),
closed = c( TRUE, T ),
type = "Z"
)
Height_cats_type1 <- Intervals(
matrix(c(100, 120, 121, 140, 141, 186), ncol = 2, byrow = TRUE ),
closed = c( TRUE, T ),
type = "Z"
)
Height_cats_type2 <- Intervals(
matrix(c(110, 125, 126, 145, 146, 190), ncol = 2, byrow = TRUE ),
closed = c( TRUE, T ),
type = "Z"
)
Weight_cats_type1 <- Intervals(
matrix(c(50, 60, 61, 78, 79, 85), ncol = 2, byrow = TRUE ),
closed = c( TRUE, T ),
type = "Z"
)
Weight_cats_type2 <- Intervals(
matrix(c(55, 75, 76, 90, 91, 100), ncol = 2, byrow = TRUE ),
closed = c( TRUE, T ),
type = "Z"
)
#now format data
# first for age
library(data.table)
PEOPLE1 <- data.table(People)
PEOPLE1[, A1:=Age]
I_age_1 <- data.table(cbind(data.frame(AGE_cats_type1), idX=1:3, idY=0))
I_age_2 <- data.table(cbind(data.frame(AGE_cats_type2), idX=0, idY=1:3))
setkey(I_age_1, X1, X2)
setkey(I_age_2, X1, X2)
PEOPLE2 <- data.frame(rbind(
foverlaps(PEOPLE1, I_age_1, by.x=c("Age", "A1"), nomatch=0),
foverlaps(PEOPLE1, I_age_2, by.x=c("Age", "A1"), nomatch=0)))
####################################################
# second iteration for height
PEOPLE3 <- data.table(PEOPLE2)
PEOPLE3[, B1:=Height]
I_height_1 <- data.table(cbind(data.frame(Height_cats_type1), idXa=1:3, idYa=0))
I_height_2 <- data.table(cbind(data.frame(Height_cats_type2), idXa=0, idYa=1:3))
setkey(I_height_1, X1, X2)
setkey(I_height_2, X1, X2)
PEOPLE4 <- data.frame(rbind(
foverlaps(PEOPLE3, I_height_1, by.x=c("Height", "B1"), nomatch=0),
foverlaps(PEOPLE3, I_height_1, by.x=c("Height", "B1"), nomatch=0)))
################################################
# third iteration
PEOPLE5 <- data.table(PEOPLE4)
PEOPLE5[, C1:=Weight]
I_weight_1 <- data.table(cbind(data.frame(Weight_cats_type1), idXb=1:3, idYb=0))
I_weight_2 <- data.table(cbind(data.frame(Weight_cats_type2), idXb=0, idYb=1:3))
setkey(I_weight_1, X1, X2)
setkey(I_weight_2, X1, X2)
PEOPLE6 <- data.frame(rbind(
foverlaps(PEOPLE5, I_weight_1, by.x=c("Height", "B1"), nomatch=0),
foverlaps(PEOPLE5, I_weight_2, by.x=c("Height", "B1"), nomatch=0)))
但在 PEOPLE6 中出现错误。
Error in setcolorder(ans, c(xcols1, ycols, xcols2)) :
neworder is length 16 but x has 18 columns.
当我查看 PEOPLE4 时,我们看到 idX idY idxA 和 idyA 是 Age_cats_type1、Age_cats_type2、Height_cat_Type1 和 Height_cat_Type2 值
【问题讨论】:
-
到目前为止你尝试过什么代码?
-
您需要提供更多reproducible problem。请首先列出您在此问题上尝试过的实际代码。由于您引用了错误,因此阅读“第三次迭代出现错误”对我们没有帮助; 什么错误?