提问人：JKO 提问时间：10/28/2020 最后编辑：mhovdJKO 更新时间：10/28/2020 访问量：80
您如何（简单地）将函数应用于 R 中不同长度的多个子集？[复制]
How do you (simply) apply a function to mutliple sub-sets of differing lengths in R? [duplicate]
问：
我需要将函数应用于列中不同长度的多个数据子集，并生成一个新的数据框，其中包括输出及其关联的元数据。
如何在不求助于 for 循环的情况下做到这一点？ 似乎是一个很好的起点，但我在语法上很挣扎。tapply()
例如，我有这样的东西：
block plot id species type response
1 1 1 w a 1.5
1 1 2 w a 1
1 1 3 w a 2
1 1 4 w a 1.5
1 2 5 x a 5
1 2 6 x a 6
1 2 7 x a 7
1 3 8 y b 10
1 3 9 y b 11
1 3 10 y b 9
1 4 11 z b 1
1 4 12 z b 3
1 4 13 z b 2
2 5 14 w a 0.5
2 5 15 w a 1
2 5 16 w a 1.5
2 6 17 x a 3
2 6 18 x a 2
2 6 19 x a 4
2 7 20 y b 13
2 7 21 y b 12
2 7 22 y b 14
2 8 23 z b 2
2 8 24 z b 3
2 8 25 z b 4
2 8 26 z b 2
2 8 27 z b 4
我想制作这样的东西：
block plot species type mean.response
1 1 w a 1.5
1 2 x a 6
1 3 y b 10
1 4 z b 2
2 5 w a 1
2 6 x a 3
2 7 y b 13
2 8 z b 3
答：
3赞
Duck
10/28/2020
#1
试试这个。可用于设置分组变量，然后计算预期变量。这里使用以下代码：group_by()summarise()dplyr
library(dplyr)
#Code
newdf <- df %>% group_by(block,plot,species,type) %>% summarise(Mean=mean(response,na.rm=T))
输出：
# A tibble: 8 x 5
# Groups: block, plot, species [8]
block plot species type Mean
<int> <int> <chr> <chr> <dbl>
1 1 1 w a 1.5
2 1 2 x a 6
3 1 3 y b 10
4 1 4 z b 2
5 2 5 w a 1
6 2 6 x a 3
7 2 7 y b 13
8 2 8 z b 3
或者使用 （ 用于省略聚合中的变量）：base R-3id
#Base R
newdf <- aggregate(response~.,data=df[,-3],mean,na.rm=T)
输出：
block plot species type response
1 1 1 w a 1.5
2 2 5 w a 1.0
3 1 2 x a 6.0
4 2 6 x a 3.0
5 1 3 y b 10.0
6 2 7 y b 13.0
7 1 4 z b 2.0
8 2 8 z b 3.0
使用的一些数据：
#Data
df <- structure(list(block = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L), plot = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L,
4L, 4L, 5L, 5L, 5L, 6L, 6L, 6L, 7L, 7L, 7L, 8L, 8L, 8L, 8L, 8L
), id = 1:27, species = c("w", "w", "w", "w", "x", "x", "x",
"y", "y", "y", "z", "z", "z", "w", "w", "w", "x", "x", "x", "y",
"y", "y", "z", "z", "z", "z", "z"), type = c("a", "a", "a", "a",
"a", "a", "a", "b", "b", "b", "b", "b", "b", "a", "a", "a", "a",
"a", "a", "b", "b", "b", "b", "b", "b", "b", "b"), response = c(1.5,
1, 2, 1.5, 5, 6, 7, 10, 11, 9, 1, 3, 2, 0.5, 1, 1.5, 3, 2, 4,
13, 12, 14, 2, 3, 4, 2, 4)), class = "data.frame", row.names = c(NA,
-27L))
2赞
G. Grothendieck
10/28/2020
#2
在末尾的注释中可重复地给出输入的情况下，使用以下任何一种：dd
# 1. aggregate.formula - base R
# Can use just response on left hand side if header doesn't matter.
aggregate(cbind(mean.response = response) ~ block + plot + species + type, dd, mean)
# 2. aggregate.default - base R
v <- c("block", "plot", "species", "type")
aggregate(list(mean.response = dd$response), dd[v], mean)
# 3. sqldf
library(sqldf)
sqldf("select block, plot, species, type, avg(response) as [mean.response]
from dd group by 1, 2, 3, 4")
# 4. data.table
library(data.table)
v <- c("block", "plot", "species", "type")
as.data.table(dd)[, .(mean.response = mean(response)), by = v]
# 5. doBy - last column of output will be labelled response.mean
library(doBy)
summaryBy(response ~ block + plot + species + type, dd)
注意
可重复形式的输入：
dd <- structure(list(block = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L), plot = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L,
4L, 4L, 5L, 5L, 5L, 6L, 6L, 6L, 7L, 7L, 7L, 8L, 8L, 8L, 8L, 8L
), id = 1:27, species = c("w", "w", "w", "w", "x", "x", "x",
"y", "y", "y", "z", "z", "z", "w", "w", "w", "x", "x", "x", "y",
"y", "y", "z", "z", "z", "z", "z"), type = c("a", "a", "a", "a",
"a", "a", "a", "b", "b", "b", "b", "b", "b", "a", "a", "a", "a",
"a", "a", "b", "b", "b", "b", "b", "b", "b", "b"), response = c(1.5,
1, 2, 1.5, 5, 6, 7, 10, 11, 9, 1, 3, 2, 0.5, 1, 1.5, 3, 2, 4,
13, 12, 14, 2, 3, 4, 2, 4)), class = "data.frame", row.names = c(NA,
-27L))
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