Question

dA有这样的数据表

id   group    startPoints       endPoints
1    A        4, 20, 50, 63,   8, 25, 60, 78
1    A        120, 300,        231, 332
1    B        500,             550
1    B        650, 800         700, 820
1    C        830, 900, 950    850, 920, 970

我想要获得的是获得特定组中长度（EndPoint - StartPoint）的SUM / MEAN /等，但是无法使用sapply

我的目标是获得表单的结果：

Group    SUM 
A        177
B        120
C        60

我正在尝试结合两件事

 lengths <- strsplit(as.character(table$endPoints), ",", fixed=TRUE)

和

y <- factor(table$group)
tapply(lengths, y, sum)

但是我被困住了，无法让它发挥作用。

添加样本数据

structure(list(id = c(1L, 1L, 1L, 1L, 1L), group = structure(c(1L, 
1L, 2L, 2L, 3L), .Label = c("A", "B", "C"), class = "factor"), 
startPoints = structure(c(2L, 1L, 3L, 4L, 5L), .Label = c("120,300,", 
"4,20,50,63,", "500,", "650,800,", "830,900,950,"), class = "factor"), 
endPoints = structure(c(4L, 1L, 2L, 3L, 5L), .Label = c("231,332,", 
"550,", "700,820,", "8,25,60,78", "850,920,970,"), class = "factor")), 
.Names = c("id", "group", "startPoints", "endPoints"), class = "data.frame", 
row.names = c(NA, -5L))

Answer 1

根据您的要求，这与{{1}}完全无关，但这是使用我的“splitstackshape”软件包中的sapply的一种方法。

首先，将数据拆分为半长格式：

concat.split.multiple

计算“endPoints”和“startPoints”之间的差异：

library(splitstackshape)
mydf2 <- concat.split.multiple(mydf, split.cols = c("startPoints", "endPoints"), 
                               seps = ",", direction = "long")

使用mydf2$diffs <- mydf2$endPoints - mydf2$startPoints head(mydf2) # id group .id time startPoints endPoints diffs # 1 1 A 1 1 4 8 4 # 2 1 A 2 1 120 231 111 # 3 1 B 1 1 500 550 50 # 4 1 B 2 1 650 700 50 # 5 1 C 1 1 830 850 20 # 6 1 A 1 2 20 25 5（或aggregate，或data.table或您最喜欢的聚合函数来计算您想要的任何内容。

tapply

Answer 2

或者更多'手动'，如果你的数据框是xx，那么分割的endPoints成为单独的元素，找出每行的长度

endPoints = strsplit(as.character(xx$endPoints), ",", fixed=TRUE)
startPoints = strsplit(as.character(xx$startPoints), ",", fixed=TRUE)
len = sapply(endPoints, length)

使用长度扩展原始数据框，取消列出以前压缩的元素

yy = cbind(xx[rep(seq_len(nrow(xx)), len), c("id", "group")], 
              startPoints=as.integer(unlist(startPoints)), 
              endPoints=as.integer(unlist(endPoints)))

之后aggregate是你的朋友。

aggregate(endPoints - startPoints ~ group, yy, sum)

使用sapply聚合列逗号分隔值

2 个答案: