[英]R — How can I calculate group means for a list of data frames, using a different subset condition to calculate each mean?
我有一个包含三个数据帧的列表,并且想生成另一个包含三个数据帧的列表,它们的行由分组变量(g1)的每个值和g1变量的六个变量的均值组成。 唯一的问题是,我只想在对应的虚拟变量的值等于1时才计算三个连续变量的均值。
可重现的示例:
a <- data.frame(c("fj","fj","fj","a","fj","a","g","g","g","g"),c(1,1,1,1,0,0,0,1,0,0),c(0,0,1,0,1,0,0,1,0,1),c(0,0,0,1,0,0,1,1,0,0),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)))
b <- data.frame(c("fj","a","fj","a","fj","fj","fj","g","g","g"),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)))
c <- data.frame(c("fj","fj","fj","a","fj","a","g","g","g","g"),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)))
u <- list(a,b,c)
u <- lapply(u, setNames, nm = c('g1','dummy1','dummy2','dummy3','contin1','contin2','contin3'))
u[[1]]
> u
[[1]]
g1 dummy1 dummy2 dummy3 contin1 contin2 contin3
1 fj 1 0 0 199 18 61
2 fj 1 0 0 91 158 28
3 fj 1 1 0 147 67 190
4 a 1 0 1 181 105 22
5 fj 0 1 0 14 16 156
6 a 0 0 0 178 14 98
7 g 0 0 1 116 97 30
8 g 1 1 1 48 31 144
9 g 0 0 0 60 21 112
10 g 0 1 0 95 145 199
我想仅在dummy1 = 1时计算contin1的平均值,仅在dummy2 = 1时才计算contin2的平均值,仅在dummy3 = 1时才计算contin3的平均值
第一个列表的输出I WANT:
> rates
[[1]]
x[, 1] V1 V2 V3 x[, 1] x[, 6] x[, 1] x[, 7] x[, 1] x[, 8]
1 a 0.50 0.0 0.5 a 181 a NA a 22
2 fj 0.75 0.5 0.0 fj 145.67 fj 41.5 fj NA
3 g 0.25 0.5 0.5 g 48 g 88 g 87
我试过的
rates <- lapply(u, function(x) {
cbind(aggregate(cbind(x[,2],x[,3],x[,4]) ~ x[,1], FUN = mean, na.action = NULL),
aggregate(x[,6] ~ x[,1], FUN = mean, na.action = NULL, subset = (x[,2] == 1)),
aggregate(x[,7] ~ x[,1], FUN = mean, na.action = NULL, subset = (x[,3] == 1)),
aggregate(x[,8] ~ x[,1], FUN = mean, na.action = NULL, subset = (x[,4] == 1)))
})
Error in data.frame(..., check.names = FALSE) :
arguments imply differing number of rows: 3, 2
我知道此错误来自cbind,因为每当您尝试对具有不同行数的对象进行绑定时,cbind都会失败。 (x [,6]列有三行,而x [,7]和x [,8]有两行。)我想我希望有某种方法可以使合计为每个分组变量保留一行,这意味着我将拥有相同数量的行,并且cbind可以工作。 也许根据R文档无法做到这一点?:“结果中将忽略任何by变量中缺少值的行。”
我已经阅读了有关汇总的文档。 以下两个帖子解决了类似的问题,但没有使用数据的不同子集来计算均值。
R:计算一组子集的 均值,并从R中的数据帧列表中 计算 均值
任何建议将不胜感激。
如果您已安装dplyr,则以下代码似乎可以解决您的问题。
library(dplyr)
set.seed(1234)
a <- data.frame(c("fj","fj","fj","a","fj","a","g","g","g","g"),c(1,1,1,1,0,0,0,1,0,0),c(0,0,1,0,1,0,0,1,0,1),c(0,0,0,1,0,0,1,1,0,0),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)))
b <- data.frame(c("fj","a","fj","a","fj","fj","fj","g","g","g"),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)))
c <- data.frame(c("fj","fj","fj","a","fj","a","g","g","g","g"),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)))
u <- list(a,b,c)
u <- lapply(u, setNames, nm = c('g1','dummy1','dummy2','dummy3','contin1','contin2','contin3'))
rates <- lapply(u, function(x)
x %>%
mutate( contin1_ = ifelse(dummy1==1, contin1, NA) ) %>%
mutate( contin2_ = ifelse(dummy2==1, contin2, NA) ) %>%
mutate( contin3_ = ifelse(dummy3==1, contin3, NA) ) %>%
group_by(g1) %>%
summarize(
V1 = mean(dummy1, na.rm=TRUE),
V2 = mean(dummy2, na.rm=TRUE),
V3 = mean(dummy3, na.rm=TRUE),
mean1 = mean(contin1_, na.rm=TRUE),
mean2 = mean(contin2_, na.rm=TRUE),
mean3 = mean(contin3_, na.rm=TRUE)
)
)
print(rates[[1]])
这给了我这个:
Source: local data frame [3 x 7]
g1 V1 V2 V3 mean1 mean2 mean3
1 a 0.50 0.0 0.5 128.00000 NaN 17
2 fj 0.75 0.5 0.0 94.66667 64 NaN
3 g 0.25 0.5 0.5 54.00000 57 146
我得到的数字似乎是正确的,并且NA在所有正确的位置。 不幸的是,您的示例不能完全重现,因为您没有指定生成随机变量的种子,因此我的runif给我的值与您的不同。
另一种选择是将格式从“宽”更改为“长”,并在获得“平均值”值后重新转换回“宽”。 对于多值列,这是现在可能melt
, dcast
从的开发人员版本data.table
即v1.9.5
。 可以从here
安装。 (使用与@akhmed帖子相同的数据集)。
我们可以melt
通过指定列的索引中的(“虚设”和“CONTIN”)的列表(“U”)内数据集measure.vars
为列表。 通过将值指定为从long
到wide
, dcast
按``g1''分组的``dummy''和``contin''列的平均值和``variable''(从``melt''创建)的变量.vars为``dummyMean''和``continMean'' 。
res <- lapply(u, function(x) {
x1 <- melt(setDT(x), measure.vars=list(2:4,5:7),
value.name=c('dummy', 'contin'))
x2 <- x1[, list(dummyMean = mean(dummy, na.rm=TRUE),
continMean = mean(contin[dummy==1], na.rm=TRUE)),
by=list(g1, variable)]
dcast(x2, g1~variable, value.var=c('dummyMean', 'continMean'))})
res[[1]]
# g1 1_dummyMean 2_dummyMean 3_dummyMean 1_continMean 2_continMean
#1: a 0.50 0.0 0.5 128.00000 NaN
#2: fj 0.75 0.5 0.0 94.66667 64
#3: g 0.25 0.5 0.5 54.00000 57
# 3_continMean
#1: 17
#2: NaN
#3: 146
或使用Map
的base R
选项。 创建了函数“ fdummy”,“ fcontin”以将“ dummy”和“ contin”列作为子集。 遍历'u'( lapply(...)
)。 使用Map
获得“虚拟”和“CONTIN”,由“G1”列分组的相应列,得到mean
“虚拟”的,并mean
使用带有“虚拟== 1” CONTIN“列tapply
, cbind
结果。
fdummy <- function(x) x[grep('dummy', names(x))]
fcontin <- function(x) x[grep('contin', names(x))]
res2 <- lapply(u, function(x) {
do.call(cbind.data.frame,
Map(function(x,y,z) cbind(tapply(x,z, FUN=mean),
tapply(y[x==1],z[x==1], FUN=mean)),
fdummy(x), fcontin(x), x['g1']))})
lapply(res2, setNames, c(rbind(paste0('dummyMean', 1:3),
paste0('continMean',1:3))))[[1]]
# dummyMean1 continMean1 dummyMean2 continMean2 dummyMean3 continMean3
#a 0.50 128.00000 0.0 NA 0.5 17
#fj 0.75 94.66667 0.5 64 0.0 NA
#g 0.25 54.00000 0.5 57 0.5 146
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