4

我正在寻找你的帮助。我试图将一个连续变量分成两组，我举了这个例子来说明我想要做什么：

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,])

在我计算 g1 和 g2 的均值和 sd 之后：

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，我想要一个显示如下内容的表格：

Value   Mean.G1  SD.G1  Mean.G2  SD.G2  N.G1 N.G2
20        62.61    26.8   6.21     3.55  13   37

但是这个表不仅适用于 20，我想为具有不同值的向量构建该表，例如一个具有 10 个元素的向量，从 20 开始，并且在步骤 20 中增加，像这样的向量v=c(10,30,50,70,90,110,130,150,170,190)。我等待这个问题是明确的。谢谢

4

4 回答 4

2

第一步可以更经济地完成，结果与以下相同：

g1=x[x$v1>20,]
g2=x[x$v1<=20,] # since "[" would have returned a dataframe

但是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

如果您想在多个位置进行剪切，请使用剪切功能来拆分和识别组：

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meltreshape2 包中的函数很有用，我注意到中断向量需要一个最低参数来拾取低于 10 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
于 2013-03-17T21:34:11.307 回答
2

您可以summary在这里简单地使用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
于 2013-03-17T20:53:36.787 回答
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
于 2013-03-17T22:16:00.070 回答
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
于 2013-03-17T20:35:12.533 回答