dplyr sample_n其中n是分组变量的值 [英] dplyr sample_n where n is the value of a grouped variable
问题描述
dplyr :: sample_n
从每个组的数据框架中提取行。我想使用每个组中分组变量 NDG
的值作为从每个组中提取的行数。 > dg.tmp< - structure(list(Gene = c(CAMK1,GHRL,TIMP4,CAMK1,GHRL,
TIMP4,ARL8B,ARPC4 SEC13,ARL8B,ARPC4,SEC13,
),GLB = c(3,3,3,3,3,10,10,10,10,10,10) NDG = c(1,
1,1,2,2,2,1,1,2,2,2,2)),class = c(tbl_df,tbl,
data.frame),row.names = c(NA,-12L),.Names = c(Gene,GLB,
NDG))
> ; dg< - dg.tmp%>%
dplyr :: group_by(GLB,NDG)
> dg
资料来源:本地数据框架[12 x 3]
组:GLB,NDG
基因GLB NDG
1 A4GNT 3 1
2 ABTB1 3 1
3 AHSG 3 1
4 A4GNT 3 2
5 ABTB1 3 2
6 AHSG 3 2
7 AADAC 10 1
8 ABHD14B 10 1
9 ACVR2B 10 1
10 AADAC 10 2
11 ABHD14B 10 2
12 ACVR2B 10 2
例如,假设正确的随机选择,我想要代码
> dg%>%dplyr :: sample_n(NDG)
输出:
资料来源:本地数据框架[6 x 3]
组:GLB,NDG
基因GLB NDG
1 A4GNT 3 1
2 A4GNT 3 2
3 ABTB1 3 2
4 AADAC 10 1
5 AADAC 10 2
6 ABHD14B 10 2
但是,它出现以下错误:
eval(expr,envir,enclosure)中的错误:未找到对象'NDG'
通过比较,当我使用代码
<$ p $时,
dplyr :: slice
给出了正确的输出p> > dg%>%dplyr :: slice(1:unique(NDG))
在这种情况下,使用独特
稍微黑客,但代码
> dg%>%dplyr :: slice(1:NDG)
返回以下警告消息
警告消息:
1:在slice_impl(.data,dots)中:
数字表达式有3个元素:只有第一个使用
2:在slice_impl(.data,dots)中:
数字表达式有3个元素:只有第一个使用
3:在slice_impl(.data,dots)中:
数字表达式有3个元素:只有第一个使用
4:在slice_impl(.data,dots)中:
数字表达式有3个元素:只有第一个使用
由于 NDG
正在评估(在适当的环境中)为 c(1,1,1)
或 c(2,2,2)
,因此 1: NDG
返回上述警告。
关于为什么我收到错误,我知道代码Hadley用于方法sample_n.grouped_df是
sample_n.grouped_df< - function(tbl,size,replace = FALSE,weight = NULL,
.env = parent.frame()){
assert_that(is.numeric(size),length(size)== 1,size> = 0 )
weight< - substitute(weight)
index< - attr(tbl,indices)
samples< - lapply(index,sample_group,frac = FALSE ,
tbl = tbl,size = size,replace = replace,weight = weight,.env = .env)
idx< - unlist(sampled)+ 1
grouping_df (tbl [idx,,drop = FALSE],vars = groups(tbl))
}
可以在相关的 Github页面中找到。因此,我得到错误,因为 sample_n.grouped_df
找不到变量 NGD
,因为它看起来不正确。 / p>
因此,在 dg $ c $上使用
sample_n
的整洁方法c>获取
资料来源:本地数据框架[6 x 3]
组:GLB,NDG
基因GLB NDG
1 A4GNT 3 1
2 A4GNT 3 2
3 ABTB1 3 2
4 AADAC 10 1
5 AADAC 10 2
6 ABHD14B 10 2
通过对每个组使用随机抽样?
一个可能的答案,但我不相信这是最佳答案:使用 dplyr ::排列数据框的行sample_frac
(和一小部分1),然后切片所需的行数:
> set.seed(1)
> dg%>%
dplyr :: sample_frac(1)%>%
dplyr :: slice(1:unique(NDG))
这给出了正确的输出。
来源:本地数据框[6 x 3]
组:GLB,NDG
基因GLB NDG
1 A4GNT 3 1
2 AHSG 3 2
3 A4GNT 3 2
4 ACVR2B 10 1
5 AADAC 10 2
6 ACVR2B 10 2
我想我可以写一个函数,如果需要,可以一行一行。
I have the following grouped data frame, and I would like to use the function dplyr::sample_n
to extract rows from this data frame for each group. I want to use the value of the grouped variable NDG
in each group as the number of rows to extract from each group.
> dg.tmp <- structure(list(Gene = c("CAMK1", "GHRL", "TIMP4", "CAMK1", "GHRL",
"TIMP4", "ARL8B", "ARPC4", "SEC13", "ARL8B", "ARPC4", "SEC13"
), GLB = c(3, 3, 3, 3, 3, 3, 10, 10, 10, 10, 10, 10), NDG = c(1,
1, 1, 2, 2, 2, 1, 1, 1, 2, 2, 2)), class = c("tbl_df", "tbl",
"data.frame"), row.names = c(NA, -12L), .Names = c("Gene", "GLB",
"NDG"))
> dg <- dg.tmp %>%
dplyr::group_by(GLB,NDG)
> dg
Source: local data frame [12 x 3]
Groups: GLB, NDG
Gene GLB NDG
1 A4GNT 3 1
2 ABTB1 3 1
3 AHSG 3 1
4 A4GNT 3 2
5 ABTB1 3 2
6 AHSG 3 2
7 AADAC 10 1
8 ABHD14B 10 1
9 ACVR2B 10 1
10 AADAC 10 2
11 ABHD14B 10 2
12 ACVR2B 10 2
For example, assuming the correct random selection, I want the code
> dg %>% dplyr::sample_n(NDG)
to output:
Source: local data frame [6 x 3]
Groups: GLB, NDG
Gene GLB NDG
1 A4GNT 3 1
2 A4GNT 3 2
3 ABTB1 3 2
4 AADAC 10 1
5 AADAC 10 2
6 ABHD14B 10 2
However, it gives the following error:
Error in eval(expr, envir, enclos) : object 'NDG' not found
By way of comparison, dplyr::slice
gives the correct output when I use the code
> dg %>% dplyr::slice(1:unique(NDG))
It is slightly hackish using unique
in this context, however, the code
> dg %>% dplyr::slice(1:NDG)
returns the following warning messages
Warning messages:
1: In slice_impl(.data, dots) :
numerical expression has 3 elements: only the first used
2: In slice_impl(.data, dots) :
numerical expression has 3 elements: only the first used
3: In slice_impl(.data, dots) :
numerical expression has 3 elements: only the first used
4: In slice_impl(.data, dots) :
numerical expression has 3 elements: only the first used
clearly because NDG
is being evaluated (in the appropriate environment) as c(1,1,1)
or c(2,2,2)
, and hence 1:NDG
returns the above warning.
Regarding why I obtain the error, I know that the code Hadley uses for the method sample_n.grouped_df is
sample_n.grouped_df <- function(tbl, size, replace = FALSE, weight = NULL,
.env = parent.frame()) {
assert_that(is.numeric(size), length(size) == 1, size >= 0)
weight <- substitute(weight)
index <- attr(tbl, "indices")
sampled <- lapply(index, sample_group, frac = FALSE,
tbl = tbl, size = size, replace = replace, weight = weight, .env = .env)
idx <- unlist(sampled) + 1
grouped_df(tbl[idx, , drop = FALSE], vars = groups(tbl))
}
which can be found on the relevant Github page. Thus I obtain the error because sample_n.grouped_df
cannot find the variable NGD
because it's not looking in the correct environment.
Consequently, is there a neat way of using sample_n
on dg
to obtain
Source: local data frame [6 x 3]
Groups: GLB, NDG
Gene GLB NDG
1 A4GNT 3 1
2 A4GNT 3 2
3 ABTB1 3 2
4 AADAC 10 1
5 AADAC 10 2
6 ABHD14B 10 2
by using random sampling on each group?
One possible answer, but I'm not convinced it's the optimal answer: permute the rows of the data frame with dplyr::sample_frac
(and a fraction of 1), then slice the required number of rows:
> set.seed(1)
> dg %>%
dplyr::sample_frac(1) %>%
dplyr::slice(1:unique(NDG))
This gives the correct output.
Source: local data frame [6 x 3]
Groups: GLB, NDG
Gene GLB NDG
1 A4GNT 3 1
2 AHSG 3 2
3 A4GNT 3 2
4 ACVR2B 10 1
5 AADAC 10 2
6 ACVR2B 10 2
And I suppose I can just write a function to do this in one line if necessary.
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