我一直试图从摘要列表中提取数据,并保留列指示符。
下面是数据的一个示例
data = data.frame( Var1 = c("Esp1"), Var2 = c("Tra1", "Tra2", "Tra3"))
data$New[[1]] <- c(list(data.frame( P.value =runif(1, 0.03, 0.08) , DF =runif(1, 1, 4) , ChisQ =runif(1, 0.03, 0.08) )),list(data.frame(factor = c("z"), value= sample(100,1))) , list(data.frame(factor = c(1:4), value= sample(100,4))) , list(data.frame( Group = c("A"), row.names =c("Control", "X1", "X2", "X3", "X4"), Value =sample(100, size=5, replace = TRUE))))
data$New[[2]] <- c(list(data.frame( P.value =runif(1, 0.03, 0.08) , DF =runif(1, 1, 4) , ChisQ =runif(1, 0.03, 0.08) )),list(data.frame(factor = c("z"), value= sample(100,1))) , list(data.frame(factor = c(1:4), value= sample(100,4))) , list(data.frame( Group = c("A"), row.names =c("Control", "X1", "X2", "X3", "X4"), Value =sample(100, size=5, replace = TRUE))))
data$New[[3]] <- c(list(data.frame( P.value =runif(1, 0.03, 0.08) , DF =runif(1, 1, 4) , ChisQ =runif(1, 0.03, 0.08) )),list(data.frame(factor = c("z"), value= sample(100,1))) , list(data.frame(factor = c(1:4), value= sample(100,4))) , list(data.frame( Group = c("A"), row.names =c("Control", "X1", "X2", "X3", "X4"), Value =sample(100, size=5, replace = TRUE))))
names(data$New[[1]]) <- c("Statistics","xx1","xx2", "groups")
names(data$New[[2]]) <- c("Statistics","xx1","xx2", "groups")
names(data$New[[3]]) <- c("Statistics","xx1","xx2", "groups")我想从每个结果列表中提取,只提取组(5行)和统计(1行)选项卡,并将它们与分别为Var 1和Var2列的值放在一个表中。在选项卡组选项卡中,row.names指示用于分析的处理方法。
我试着使用broom::tidy (它适用于其他结果列表),但是这个列表分发失败了。
到目前为止,我已经能够从提取的组及其相关的行名创建表,但是无法正确地设置正确的Var1和Var2值。
data.1 <- lapply(data[[3]], function(x) x$groups)
data.2 <- lapply(data.1, function(x) { x$Treatment <-rownames(x);return(x)})
data.group<- do.call(rbind.data.frame, data.2)
rownames(data.group) <- 1:nrow(data.group)这就是我有多远
> data.group
Group Value Treatment
1 A 24 Control
2 A 96 X1
3 A 76 X2
4 A 26 X3
5 A 10 X4
6 A 58 Control
7 A 33 X1
8 A 30 X2
9 A 54 X3
10 A 48 X4
11 A 66 Control
12 A 80 X1
13 A 97 X2
14 A 86 X3
15 A 86 X4此行不工作,无法使用于读取列Var1和Var2
data.2.1 <- lapply(data.2, function(x) { x$Var1 <-unlist(data$Var1[[(x)]]) ;return(x)})
data.2.2 <- lapply(data.2, function(x) { x$Var2 <-data[[2]][[x]] ;return(x)})这就是我希望表输出的方式
> data.group
Var1 Var2 Treatment Value Group P.value Df
1 Esp1 Tra1 Control 70 A 0.0730726366001181 0.0566315333195962
2 Esp1 Tra1 X1 27 A
3 Esp1 Tra1 X2 3 A
4 Esp1 Tra1 X3 16 A
5 Esp1 Tra1 X4 58 A
6 Esp1 Tra2 Control 2 A 0.0669188804645091 0.043313137262594
7 Esp1 Tra2 X1 58 A
8 Esp1 Tra2 X2 87 A
9 Esp1 Tra2 X3 12 A
10 Esp1 Tra2 X4 23 A
11 Esp1 Tra3 Control 58 A 0.0698359214654192 0.0380288420431316
12 Esp1 Tra3 X1 80 A
13 Esp1 Tra3 X2 44 A
14 Esp1 Tra3 X3 100 A
15 Esp1 Tra3 X4 78 A
ChisQ
1 0.0551552523346618
2
3
4
5
6 0.0415172106772661
7
8
9
10
11 0.0434505424182862
12
13
14
15
> 谢谢你的回复!!
发布于 2020-02-05 03:12:10
由于Statistics和groups是包含不相关信息的数据格式,所以我建议您将它们保存在列表中,并只选择感兴趣的数据。
data$New <- lapply(data$New,function(x)
list(Statistics = x$Statistics, groups = x$groups))
data$New
#[[1]]
#[[1]]$Statistics
# P.value DF ChisQ
#1 0.0747 2.22 0.0345
#[[1]]$groups
# Group Value
#Control A 98
#X1 A 76
#X2 A 71
#X3 A 62
#X4 A 25
#[[2]]
#[[2]]$Statistics
# P.value DF ChisQ
#1 0.074 3.71 0.0781
#[[2]]$groups
# Group Value
#Control A 31
#X1 A 92
#...
#....对于更新的预期输出,我们可以这样做
list_df <- lapply(data$New,function(x) data.frame(Control = rownames(x$groups),
Value = x$groups$Value, Group = x$groups$Group, P.value = x$Statistics$P.value,
DF = x$Statistics$DF, ChisQ = x$Statistics$ChisQ))
new_df <- data[rep(seq_len(nrow(data)), sapply(list_df, nrow)), ]
new_df$New <- NULL
cbind(new_df, do.call(rbind, list_df))或类似但使用tidyverse
data$New <- purrr::map(data$New,function(x) data.frame(Control = rownames(x$groups),
Value = x$groups$Value, Group = x$groups$Group, P.value = x$Statistics$P.value,
DF = x$Statistics$DF, ChisQ = x$Statistics$ChisQ))
data %>% tidyr::unnest(New)发布于 2020-02-05 03:33:08
我们可以使用来自map的purrr
library(purrr)
library(dplyr)
data <- data %>%
mutate(New = map(New, ~ list(Statistics = .x$Statistics, groups = .x$groups)))
data$New
#[[1]]
#[[1]]$Statistics
# P.value DF ChisQ
#1 0.05901864 2.223526 0.07536408
#[[1]]$groups
# Group Value
#Control A 62
#X1 A 49
#X2 A 69
#X3 A 15
#X4 A 88
#[[2]]
#[[2]]$Statistics
# P.value DF ChisQ
#1 0.06802287 1.506049 0.06263245
#[[2]]$groups
# Group Value
# ...https://stackoverflow.com/questions/60068301
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