我有一个包含以下信息的数据集：
=ARRAYFORMULA({"Date", "Name", "Type", "Value"; QUERY({
QUERY({B3:I, IF(F3:F<>"", $F$2, )},
"select Col1, Col2, Col9, Col5 where Col9 is not null", 0);
QUERY({B3:I, IF(G3:G<>"", $G$2, )},
"select Col1, Col2, Col9, Col6 where Col9 is not null", 0);
QUERY({B3:I, IF(H3:H<>"", $H$2, )},
"select Col1, Col2, Col9, Col7 where Col9 is not null", 0);
QUERY({B3:I, IF(I3:I<>"", $I$2, )},
"select Col1, Col2, Col9, Col8 where Col9 is not null", 0)},
"select * order by Col1 asc", 0)})
如果UniqueNumber的值> 0，我想对从第1行到UniqueNumber的每个主题的dplyr值求和，并计算平均值。因此，对于主题001，总和= 2，平均值= .67。
Subject Value1 Value2 Value3 UniqueNumber
001 1 0 1 3
002 0 1 1 2
003 1 1 1 1
编辑：我只是想对“ UniqueNumber”列中列出的列数求和。因此，这遍历每一行，并在“ UniqueNumber”中列出的列处停止。 示例：具有主题002的第2行应将“ Value1”和“ Value2”列中的值相加，而具有主题003的第3行应仅将“ Value1”列中的值相加。
答案 0 :(得分：9)
不是整洁的爱好者/专家，但我会尝试使用长格式。然后，只需按组的行索引进行过滤，然后在单个列上运行您想要的任何功能（这样更容易）。
library(tidyr)
library(dplyr)
Data %>%
gather(variable, value, -Subject, -UniqueNumber) %>% # long format
group_by(Subject) %>% # group by Subject in order to get row counts
filter(row_number() <= UniqueNumber) %>% # filter by row index
summarise(Mean = mean(value), Total = sum(value)) %>% # do the calculations
ungroup()
## A tibble: 3 x 3
# Subject Mean Total
# <int> <dbl> <int>
# 1 1 0.667 2
# 2 2 0.5 1
# 3 3 1 1
一种非常类似的方法可以通过列名中的整数进行过滤。过滤器步骤位于group_by之前，因此它有可能提高性能（或没有提高？），但它的鲁棒性较弱，因为我假设感兴趣的列被称为"Value#"
Data %>%
gather(variable, value, -Subject, -UniqueNumber) %>% #long format
filter(as.numeric(gsub("Value", "", variable, fixed = TRUE)) <= UniqueNumber) %>% #filter
group_by(Subject) %>% # group by Subject
summarise(Mean = mean(value), Total = sum(value)) %>% # do the calculations
ungroup()
## A tibble: 3 x 3
# Subject Mean Total
# <int> <dbl> <int>
# 1 1 0.667 2
# 2 2 0.5 1
# 3 3 1 1
只是为了好玩，添加一个data.table解决方案
library(data.table)
data.table(Data) %>%
melt(id = c("Subject", "UniqueNumber")) %>%
.[as.numeric(gsub("Value", "", variable, fixed = TRUE)) <= UniqueNumber,
.(Mean = round(mean(value), 3), Total = sum(value)),
by = Subject]
# Subject Mean Total
# 1: 1 0.667 2
# 2: 2 0.500 1
# 3: 3 1.000 1
答案 1 :(得分：2)
检查此解决方案：
df %>%
gather(key, val, Value1:Value3) %>%
group_by(Subject) %>%
mutate(
Sum = sum(val[c(1:(UniqueNumber[1]))]),
Mean = mean(val[c(1:(UniqueNumber[1]))]),
) %>%
spread(key, val)
输出：
Subject UniqueNumber Sum Mean Value1 Value2 Value3
<chr> <int> <dbl> <dbl> <dbl> <dbl> <dbl>
1 001 3 2 0.667 1 0 1
2 002 2 1 0.5 0 1 1
3 003 1 1 1 1 1 1
答案 2 :(得分：1)
使用purrr::map_df（与dplyr来自同一作者的解决方案）。
library(dplyr)
library(purrr)
l_dat <- split(dat, dat$Subject) # first we need to split in a list
map_df(l_dat, function(x) {
n_cols <- x$UniqueNumber # finds the number of columns
x <- as.numeric(x[2:(n_cols+1)]) # subsets x and converts to numeric
mean(x, na.rm=T) # mean to be returned
})
# output:
# # A tibble: 1 x 3
# `1` `2` `3`
# <dbl> <dbl> <dbl>
# 1 0.667 0.5 1
另一种选择（输出格式更接近dplyr解决方案）：
map_df(l_dat, function(x) {
n_cols <- x$UniqueNumber
id <- x$Subject
x <- as.numeric(x[2:(n_cols+1)])
tibble(id=id, mean_values=mean(x, na.rm=T))
})
# # A tibble: 3 x 2
# id mean_values
# <int> <dbl>
# 1 1 0.667
# 2 2 0.5
# 3 3 1
仅作为示例，我添加了sum()然后除以length(x)-1：
map_df(l_dat, function(x) {
n_cols <- x$UniqueNumber
id <- x$Subject
x <- as.numeric(x[2:(n_cols+1)])
tibble(id=id,
mean_values=sum(x, na.rm=T)/(length(x)-1)) # change here
})
# # A tibble: 3 x 2
# id mean_values
# <int> <dbl>
# 1 1 1.
# 2 2 1.
# 3 3 Inf #beware of this case where you end up dividing by 0
数据：
tt <- "Subject Value1 Value2 Value3 UniqueNumber
001 1 0 1 3
002 0 1 1 2
003 1 1 1 1"
dat <- read.table(text=tt, header=T)
答案 3 :(得分：1)
OP可能仅对dplyr解决方案感兴趣，但出于比较目的和将来的读者使用mapply的基本R选项
cols <- grep("^Value", names(df))
cbind(df, t(mapply(function(x, y) {
if (y > 0) {
vals = as.numeric(df[x, cols[1:y]])
c(Sum = sum(vals, na.rm = TRUE), Mean = mean(vals, na.rm = TRUE))
}
else
c(0, 0)
},1:nrow(df), df$UniqueNumber)))
# Subject Value1 Value2 Value3 UniqueNumber Sum Mean
#1 1 1 0 1 3 2 0.667
#2 2 0 1 1 2 1 0.500
#3 3 1 1 1 1 1 1.000
这里，我们根据各行各自的UniqueNumber子集各行，然后如果sum的值大于0或仅返回0，则计算其为mean和UniqueNumber。
答案 4 :(得分：1)
这是使用tidyr::nest将Values列收集到列表中的另一种方法，以便我们可以使用map2遍历表。在每一行中，我们从Values列表列中选择正确的值，并分别取总和或均值。
library(tidyverse)
tbl <- read_table2(
"Subject Value1 Value2 Value3 UniqueNumber
001 1 0 1 3
002 0 1 1 2
003 1 1 1 1"
)
tbl %>%
filter(UniqueNumber > 0) %>%
nest(starts_with("Value"), .key = "Values") %>%
mutate(
sum = map2_dbl(UniqueNumber, Values, ~ sum(.y[1:.x], na.rm = TRUE)),
mean = map2_dbl(UniqueNumber, Values, ~ mean(as.numeric(.y[1:.x], na.rm = TRUE))),
)
#> # A tibble: 3 x 5
#> Subject UniqueNumber Values sum mean
#> <chr> <dbl> <list> <dbl> <dbl>
#> 1 001 3 <tibble [1 × 3]> 2 0.667
#> 2 002 2 <tibble [1 × 3]> 1 0.5
#> 3 003 1 <tibble [1 × 3]> 1 1
由reprex package（v0.2.1）于2019-02-14创建
答案 5 :(得分：1)
我认为最简单的方法是将实际上应该为NA的零设置为NA，然后在相应的列子集上使用rowSums和rowMeans。 / p>
Data[2:4][(col(dat[2:4])>dat[[5]])] <- NA
Data
# Subject Value1 Value2 Value3 UniqueNumber
# 1 1 1 0 1 3
# 2 2 0 1 NA 2
# 3 3 1 NA NA 1
library(dplyr)
Data%>%
mutate(sum = rowSums(.[2:4], na.rm = TRUE),
mean = rowMeans(.[2:4], na.rm = TRUE))
# Subject Value1 Value2 Value3 UniqueNumber sum mean
# 1 1 1 0 1 3 2 0.6666667
# 2 2 0 1 NA 2 1 0.5000000
# 3 3 1 NA NA 1 1 1.0000000
或transform(Data, sum = rowSums(Data[2:4],na.rm = TRUE), mean = rowMeans(Data[2:4],na.rm = TRUE))留在基数R中。
数据
Data <- structure(
list(Subject = 1:3,
Value1 = c(1L, 0L, 1L),
Value2 = c(0L, 1L, NA),
Value3 = c(1L, NA, NA),
UniqueNumber = c(3L, 2L, 1L)),
.Names = c("Subject","Value1", "Value2", "Value3", "UniqueNumber"),
row.names = c(NA, 3L), class = "data.frame")