Question

我想使用ddply通过多种因素来汇总来自多个变量的数据。

我有以下测试数据：

site    block   plot    rep name    weight  height  dtf
Alberta 1   2   1   A   43  139 54
Alberta 2   5   2   A   46  139 46
Alberta 4   10  3   A   49  136 54
Nunavut 1   1   1   A   49  136 59
Nunavut 2   4   2   A   51  135 50
Nunavut 3   8   3   A   52  133 56
Alberta 5   13  1   B   55  132 50
Alberta 4   12  2   B   55  125 46
Alberta 5   15  3   B   56  120 46
Nunavut 5   14  1   B   57  119 54
Nunavut 5   13  2   B   58  119 55
Nunavut 4   11  3   B   59  118 51
...

等等。

我想取变量“weight”，“height”，“dtf”，并根据“site”和“name”等因素对它们进行汇总。

我从列名称向量开始：

data.factors <- NULL
data.variables <- NULL
for(n in 1:length(data)){if(is.factor(data[[n]])){ data.factors <- c(data.factors,colnames(data[n]))} else next}
for(n in 1:length(data)){if(is.numeric(data[[n]]) || is.integer(data[[n]])){ data.variables <- c(data.variables,colnames(data[n]))} else next}

这适用于执行多个单因子anovas：

for(variables in data.variables){
for(factors in data.factors){
output1 <- aov(lm(data[[variables]]~data[[factors]]))
cat(variables)
cat(" by ")
cat(factors)
cat("\n")
print(summary(output1))
}}

但我无法与ddply合作。

for (x in data.variables){
variable.summary <- ddply(data, .(site,name), summarise,
N    = sum(!is.na(x[1])),
min = min(x[1], na.rm=TRUE),
max = max(x[1], na.rm=TRUE),
mean = mean(x[1], na.rm=TRUE),
sd   = sd(x[1], na.rm=TRUE),
se   = sd / sqrt(N)
)
print(variable.summary)
}

我得到的是以下内容：

site name N    min    max mean sd se
1  Alberta    A 1 weight weight   NA NA NA
2  Alberta    B 1 weight weight   NA NA NA
3  Alberta    C 1 weight weight   NA NA NA
4  Alberta    D 1 weight weight   NA NA NA
5  Alberta    E 1 weight weight   NA NA NA
6  Nunavut    A 1 weight weight   NA NA NA
7  Nunavut    B 1 weight weight   NA NA NA
8  Nunavut    C 1 weight weight   NA NA NA
9  Nunavut    D 1 weight weight   NA NA NA
10 Nunavut    E 1 weight weight   NA NA NA
....

如果我使用单个变量测试ddply（直接输入而不是通过“x”引用），它可以正常工作。

是否有一个技巧让函数识别引用的列ID？我已经习惯了PERL，它的$ Scalars可以在任何地方引用，并且希望R中有类似的系统。

Answer 1

ddply的继承者dplyr可以使用group_by()和summarise_each()轻松完成此操作，无需循环任何内容：

df <- data.frame(site = c("Alberta", "Alberta", "Alberta", "Nunavut", "Nunavut", "Nunavut", "Alberta", "Alberta", "Alberta", "Nunavut", "Nunavut", "Nunavut"),
                 block = c(1, 2, 4, 1, 2, 3, 5, 4, 5, 5, 5, 4),
                 plot = c(2, 5, 10, 1, 4, 8, 13, 12, 15, 14, 13, 11),
                 rep = c(1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3),
                 name = c("A", "A", "A", "A", "A", "A", "B", "B", "B", "B", "B", "B"),
                 weight = c(43, 46, 49, 49, 51, 52, 55, 55, 56, 57, 58, 59),
                 height = c(139, 139, 136, 136, 135, 133, 132, 125, 120, 119, 119, 118),
                 dtf = c(54, 46, 54, 59, 50, 56, 50, 46, 46, 54, 55, 51))

library(dplyr)

df.summary <- df %>%
  group_by(site, name) %>%
  summarise_each(funs(sum, min, max, mean, sd), weight, height, dtf)

这会产生如下数据框：

> df.summary
Source: local data frame [4 x 17]
Groups: site

     site name weight_length height_length dtf_length weight_min height_min dtf_min
1 Alberta    A             3             3          3         43        136      46
2 Alberta    B             3             3          3         55        120      46
3 Nunavut    A             3             3          3         49        133      50
4 Nunavut    B             3             3          3         57        118      51
Variables not shown: weight_max (dbl), height_max (dbl), dtf_max (dbl), weight_mean (dbl),
  height_mean (dbl), dtf_mean (dbl), weight_sd (dbl), height_sd (dbl), dtf_sd (dbl)

您可以将所需的任何功能传递给funs()内的summarise_each，因此如果您想要标准错误的列，请先将该功能设为：

se <- function(x) {
  N <- sum(!is.na(x[1]))
  return(sd / sqrt(N))
}

并通过：summarise_each(funs(sum, min, max, mean, sd, se)...)

Answer 2

尝试使用data.table：

> testdt = data.table(testdf)
> testdt[,list(meanwt=mean(weight),meanht=mean(height) ),by=list(site,name)]
      site name   meanwt   meanht
1: Alberta    A 46.00000 138.0000
2: Nunavut    A 50.66667 134.6667
3: Alberta    B 55.33333 125.6667
4: Nunavut    B 58.00000 118.6667

Max，min等可以添加到功能列表中。

R ddply循环;多种因素

2 个答案: