我正在寻求你的帮助。我试图将一个连续变量分成两组，我把这个例子放在我想要做的事情上：
x=data.frame(v1=c(1,1,2,2,3,4,5,6,9,9,11,2,4,45,67,89,1,1,5,5,5,6,6,6,9,9,9,11,11,8,8,8,51,90,40,15,30,11,8,9,9,1,5,5,100,67,78,98,34,25))
我尝试将我的连续变量分成两组，初始值为20，然后：
g1=data.frame(x[x$v1>20,])
g2=data.frame(x[x$v1<=20,])
我为g1和g2计算平均值和sd后：
mean(g1$x.x.v1...20...)=62.61538
mean(g2$x.x.v1....20...)=6.216216
sd(g1$x.x.v1...20...)=26.80963
sd(g2$x.x.v1....20...)=3.55227
length(g1$x.x.v1...20...)= 13
length(g2$x.x.v1....20...)=37
在此之后，我希望有一个表格显示如下：
Value Mean.G1 SD.G1 Mean.G2 SD.G2 N.G1 N.G2
20 62.61 26.8 6.21 3.55 13 37
但是这个表不仅仅是20我想为具有不同值的向量构建该表，例如具有10个元素的向量，并且从20开始并且在步骤op 20中增加，像这样的向量{{ 1}}。 我等这个问题很清楚。感谢
答案 0 :(得分：2)
我会使用reshape2和plyr，
library(plyr) ; library(reshape2)
v=c(10,20,30,50,70,90,110,130,150,170,190) # added 20 for checking
# create new dichotomy id variable
l1 = llply(v, function(.v) transform(x, test = x[["v1"]] <= .v))
names(l1) = v # name list elements for later reference
all = melt(l1, id=c("v1","test")) # merge data.frames together
# summarise the data by groups
results = ddply(all, c("L1","test"), summarise,
mean = mean(v1), sd=sd(v1), length=length(v1))
导致
arrange(results, as.numeric(L1))
L1 test mean sd length
1 10 FALSE 48.500000 32.505656 18
2 10 TRUE 5.343750 2.902828 32
3 20 FALSE 62.615385 26.809633 13
4 20 TRUE 6.216216 3.552270 37
5 30 FALSE 69.000000 23.870484 11
6 30 TRUE 7.307692 5.907862 39
7 50 FALSE 80.000000 17.270950 8
8 50 TRUE 9.619048 10.245647 42
9 70 FALSE 91.000000 8.717798 5
10 70 TRUE 13.088889 16.555447 45
11 90 FALSE 99.000000 1.414214 2
12 90 TRUE 17.625000 23.951747 48
13 110 TRUE 20.880000 28.456655 50
14 130 TRUE 20.880000 28.456655 50
15 150 TRUE 20.880000 28.456655 50
16 170 TRUE 20.880000 28.456655 50
17 190 TRUE 20.880000 28.456655 50
答案 1 :(得分：2)
您可以在summary
lapply
do.call(rbind,lapply( v,function(x) {
v1.inf <- summary(v1[v1<=x])
v1.sup <- summary(v1[v1>x])
m <- as.matrix(rbind(v1.inf,v1.sup))
rownames(m) <- paste(x,c('inf','sup'),sep='')
m
}))
Min. 1st Qu. Median Mean 3rd Qu. Max.
10inf 1 2.75 5.0 5.344 8.00 9
10sup 11 17.50 42.5 48.500 75.25 100
20inf 1 4.00 6.0 6.216 9.00 15
20sup 25 40.00 67.0 62.620 89.00 100
30inf 1 4.00 6.0 7.308 9.00 30
30sup 34 48.00 67.0 69.000 89.50 100
50inf 1 4.25 7.0 9.619 9.00 45
50sup 51 67.00 83.5 80.000 92.00 100
70inf 1 5.00 8.0 13.090 11.00 67
70sup 78 89.00 90.0 91.000 98.00 100
90inf 1 5.00 8.0 17.620 12.00 90
90sup 98 98.50 99.0 99.000 99.50 100
110inf 1 5.00 8.5 20.880 22.50 100
110sup NA NA NA NaN NA NA
130inf 1 5.00 8.5 20.880 22.50 100
130sup NA NA NA NaN NA NA
150inf 1 5.00 8.5 20.880 22.50 100
150sup NA NA NA NaN NA NA
170inf 1 5.00 8.5 20.880 22.50 100
170sup NA NA NA NaN NA NA
190inf 1 5.00 8.5 20.880 22.50 100
190sup NA NA NA NaN NA NA
答案 2 :(得分：2)
第一步可以更经济地完成，结果如下：
g1=x[x$v1>20,]
g2=x[x$v1<=20,] # since "[" would have returned a dataframe
但为什么不跳过这一步而是这样做：
do.call(cbind, by(x$v1, list(v1GT20 = x$v1 > 20),
function(v) c(Mean=mean(v), SD=sd(v), N=length(v)) ) )
FALSE TRUE
Mean 6.21622 62.6154
SD 3.55227 26.8096
N 37.00000 13.0000
如果您希望在各种位置获得剪切，请使用剪切功能来分割和识别组：
do.call(cbind, by(x$v1, cut( x$v1 , breaks=c(10,30,50,70,90,110,130,150,170,190) ),
function(v) c(Mean=mean(v), SD=sd(v), N=length(v)) ) )
(10,30] (30,50] (50,70] (70,90] (90,110]
Mean 16.28571 39.66667 61.6667 85.66667 99.00000
SD 7.93125 5.50757 9.2376 6.65833 1.41421
N 7.00000 3.00000 3.0000 3.00000 2.00000
如果你想要长格式，那么包reshape2中的melt函数很有用，我注意到break向量需要一个最低的参数来拾取10以下的项目：
> melt( do.call(cbind, by(x$v1,
cut( x$v1 , breaks=c(-Inf, 10,30,50,70,90,110,130,150,170,190),
include.lowest=TRUE ),
function(v) c(Mean=mean(v), SD=sd(v), N=length(v)) ) ) )
Var1 Var2 value
1 Mean [-Inf,10] 5.34375
2 SD [-Inf,10] 2.90283
3 N [-Inf,10] 32.00000
4 Mean (10,30] 16.28571
5 SD (10,30] 7.93125
6 N (10,30] 7.00000
7 Mean (30,50] 39.66667
8 SD (30,50] 5.50757
9 N (30,50] 3.00000
10 Mean (50,70] 61.66667
11 SD (50,70] 9.23760
12 N (50,70] 3.00000
13 Mean (70,90] 85.66667
14 SD (70,90] 6.65833
15 N (70,90] 3.00000
16 Mean (90,110] 99.00000
17 SD (90,110] 1.41421
18 N (90,110] 2.00000
答案 3 :(得分：2)
这是一个data.table解决方案：
require(data.table)
x.dt <- data.table(x)
rbindlist(lapply(v, function(i) {
lbls <- paste0(c(">", "<="), i)
x.dt[, grp := as.character(factor(v1 > i, levels=c(TRUE, FALSE), labels=lbls))]
x.dt[, as.list(c(v = i, mean = mean(v1),
sd = sd(v1), length = length(v1))), by = grp]
}))
# grp v mean sd length
# 1: <=10 10 5.343750 2.902828 32
# 2: >10 10 48.500000 32.505656 18
# 3: <=20 20 6.216216 3.552270 37
# 4: >20 20 62.615385 26.809633 13
# 5: <=30 30 7.307692 5.907862 39
# 6: >30 30 69.000000 23.870484 11
# 7: <=50 50 9.619048 10.245647 42
# 8: >50 50 80.000000 17.270950 8
# 9: <=70 70 13.088889 16.555447 45
# 10: >70 70 91.000000 8.717798 5
# 11: <=90 90 17.625000 23.951747 48
# 12: >90 90 99.000000 1.414214 2
# 13: <=110 110 20.880000 28.456655 50
# 14: <=130 130 20.880000 28.456655 50
# 15: <=150 150 20.880000 28.456655 50
# 16: <=170 170 20.880000 28.456655 50
# 17: <=190 190 20.880000 28.456655 50