这是我的样本数据集：
vector1 <-
data.frame(
"name" = "a",
"age" = 10,
"fruit" = c("orange", "cherry", "apple"),
"count" = c(1, 1, 1),
"tag" = c(1, 1, 2)
)
vector2 <-
data.frame(
"name" = "b",
"age" = 33,
"fruit" = c("apple", "mango"),
"count" = c(1, 1),
"tag" = c(2, 2)
)
vector3 <-
data.frame(
"name" = "c",
"age" = 58,
"fruit" = c("cherry", "apple"),
"count" = c(1, 1),
"tag" = c(1, 1)
)
list <- list(vector1, vector2, vector3)
print(list)
这是我的测试：
default <- c("cherry",
"orange",
"apple",
"mango")
for (num in 1:length(list)) {
#print(list[[num]])
list[[num]] <- rbind(
list[[num]],
data.frame(
"name" = list[[num]]$name,
"age" = list[[num]]$age,
"fruit" = setdiff(default, list[[num]]$fruit),#add missed value
"count" = 0,
"tag" = 1 #not found solutions
)
)
print(paste0("--------------", num, "--------"))
print(list)
}
#print(list)
我正在尝试在数据框中找到哪些水果遗漏，而水果是基于标记的值。例如，在第一个数据框中，有标记1和2.如果标记1的值没有默认水果，如苹果和香蕉，错过的默认水果将被添加到0到数据框。期望格式如下：
[[1]]
name age fruit count tag
1 a 10 orange 1 1
2 a 10 cherry 1 1
3 a 10 apple 1 2
4 a 10 mango 0 1
5 a 10 apple 0 1
6 a 10 mango 0 2
7 a 10 orange 0 2
8 a 10 cherry 0 2
当我检查循环的过程时，我还发现第一个循环添加芒果3次，我没有找到它无法一次添加错过的值的原因。整体输出如下：< / p>
[[1]]
name age fruit count tag
1 a 10 orange 1 1
2 a 10 cherry 1 1
3 a 10 apple 1 2
4 a 10 mango 0 1
5 a 10 mango 0 1
6 a 10 mango 0 1
[[2]]
name age fruit count tag
1 b 33 apple 1 2
2 b 33 mango 1 2
3 b 33 cherry 0 1
4 b 33 orange 0 1
[[3]]
name age fruit count tag
1 c 58 cherry 1 1
2 c 58 apple 1 1
3 c 58 orange 0 1
4 c 58 mango 0 1
有没有人帮助我并提供简单的方法或其他方法？我应该使用sqldf函数来添加0值吗？这是解决我问题的简单方法吗？
答案 0 :(得分：2)
考虑基础R方法 - lapply，expand.grid，transform，rbind，aggregate - 附加所有可能的水果和标记选项并保留最大计数。
new_list <- lapply(list, function(df) {
fruit_tag_df <- transform(expand.grid(fruit=c("apple", "cherry", "mango", "orange"),
tag=c(1,2)),
name = df$name[1],
age = df$age[1],
count = 0)
aggregate(.~name + age + fruit + tag, rbind(df, fruit_tag_df), FUN=max)
})
输出
new_list
# [[1]]
# name age fruit tag count
# 1 a 10 apple 1 0
# 2 a 10 cherry 1 1
# 3 a 10 orange 1 1
# 4 a 10 mango 1 0
# 5 a 10 apple 2 1
# 6 a 10 cherry 2 0
# 7 a 10 orange 2 0
# 8 a 10 mango 2 0
# [[2]]
# name age fruit tag count
# 1 b 33 apple 1 0
# 2 b 33 mango 1 0
# 3 b 33 cherry 1 0
# 4 b 33 orange 1 0
# 5 b 33 apple 2 1
# 6 b 33 mango 2 1
# 7 b 33 cherry 2 0
# 8 b 33 orange 2 0
# [[3]]
# name age fruit tag count
# 1 c 58 apple 1 1
# 2 c 58 cherry 1 1
# 3 c 58 mango 1 0
# 4 c 58 orange 1 0
# 5 c 58 apple 2 0
# 6 c 58 cherry 2 0
# 7 c 58 mango 2 0
# 8 c 58 orange 2 0
答案 1 :(得分：2)
OP要求完成list中的每个data.frame，以便default水果和标记1:2的所有组合都会出现在count应该是0的结果中为其他行设置为lapply()。最后，每个data.frame应至少包含 4 x 2 = 8 行。
我想提出两种不同的方法：
CJ()和来自data.table的{{1}}（交叉加入）功能返回列表。list将rbindlist()中的单独data.frames与一个大型data.table相结合，并对整个data.table应用所需的转换。lapply()和CJ() library(data.table)
lapply(lst, function(x) setDT(x)[
CJ(name = name, age = age, fruit = default, tag = 1:2, unique = TRUE),
on = .(name, age, fruit, tag)][
is.na(count), count := 0][order(-count, tag)]
)
[[1]] name age fruit count tag 1: a 10 cherry 1 1 2: a 10 orange 1 1 3: a 10 apple 1 2 4: a 10 apple 0 1 5: a 10 mango 0 1 6: a 10 cherry 0 2 7: a 10 mango 0 2 8: a 10 orange 0 2 [[2]] name age fruit count tag 1: b 33 apple 1 2 2: b 33 mango 1 2 3: b 33 apple 0 1 4: b 33 cherry 0 1 5: b 33 mango 0 1 6: b 33 orange 0 1 7: b 33 cherry 0 2 8: b 33 orange 0 2 [[3]] name age fruit count tag 1: c 58 apple 1 1 2: c 58 cherry 1 1 3: c 58 mango 0 1 4: c 58 orange 0 1 5: c 58 apple 0 2 6: c 58 cherry 0 2 7: c 58 mango 0 2 8: c 58 orange 0 2
不需要按count和tag排序，但有助于将结果与OP的预期输出进行比较。
我们可以使用一个大型data.table来代替具有相同结构的data.frames列表，其中每行的来源可以由id列标识。
事实上，OP已经提出了其他问题（"using lapply function and list in r"
和"how to loop the dataframe using sqldf?"，他在处理数据列表时请求帮助。G. Grothendieck已经建议rbind行一起。
rbindlist()函数有idcol参数，用于标识每行的来源：
library(data.table)
rbindlist(list, idcol = "df")
df name age fruit count tag 1: 1 a 10 orange 1 1 2: 1 a 10 cherry 1 1 3: 1 a 10 apple 1 2 4: 2 b 33 apple 1 2 5: 2 b 33 mango 1 2 6: 3 c 58 cherry 1 1 7: 3 c 58 apple 1 1
请注意，df包含list中的源data.frame的编号（如果list已命名，则包含列表元素的名称）。
现在，我们可以通过对df：
rbindlist(list, idcol = "df")[, .SD[
CJ(name = name, age = age, fruit = default, tag = 1:2, unique = TRUE),
on = .(name, age, fruit, tag)], by = df][
is.na(count), count := 0][order(df, -count, tag)]
df name age fruit count tag 1: 1 a 10 cherry 1 1 2: 1 a 10 orange 1 1 3: 1 a 10 apple 1 2 4: 1 a 10 apple 0 1 5: 1 a 10 mango 0 1 6: 1 a 10 cherry 0 2 7: 1 a 10 mango 0 2 8: 1 a 10 orange 0 2 9: 2 b 33 apple 1 2 10: 2 b 33 mango 1 2 11: 2 b 33 apple 0 1 12: 2 b 33 cherry 0 1 13: 2 b 33 mango 0 1 14: 2 b 33 orange 0 1 15: 2 b 33 cherry 0 2 16: 2 b 33 orange 0 2 17: 3 c 58 apple 1 1 18: 3 c 58 cherry 1 1 19: 3 c 58 mango 0 1 20: 3 c 58 orange 0 1 21: 3 c 58 apple 0 2 22: 3 c 58 cherry 0 2 23: 3 c 58 mango 0 2 24: 3 c 58 orange 0 2 df name age fruit count tag
答案 2 :(得分：1)
使用dplyr和tidyr的解决方案。我们可以使用complete展开数据框，并将填充值指定为0到count。
请注意，我将列表名称从list更改为fruit_list，因为在R中使用保留字来命名对象是一种不好的做法。另请注意，在创建示例数据框时，我设置了stringsAsFactors = FALSE，因为我不想创建因子列。最后，我使用lapply而不是for循环遍历列表元素。
library(dplyr)
library(tidyr)
fruit_list2 <- lapply(fruit_list, function(x){
x2 <- x %>%
complete(name, age, fruit = default, tag = c(1, 2), fill = list(count = 0)) %>%
select(name, age, fruit, count, tag) %>%
arrange(tag, fruit) %>%
as.data.frame()
return(x2)
})
fruit_list2
# [[1]]
# name age fruit count tag
# 1 a 10 apple 0 1
# 2 a 10 cherry 1 1
# 3 a 10 mango 0 1
# 4 a 10 orange 1 1
# 5 a 10 apple 1 2
# 6 a 10 cherry 0 2
# 7 a 10 mango 0 2
# 8 a 10 orange 0 2
#
# [[2]]
# name age fruit count tag
# 1 b 33 apple 0 1
# 2 b 33 cherry 0 1
# 3 b 33 mango 0 1
# 4 b 33 orange 0 1
# 5 b 33 apple 1 2
# 6 b 33 cherry 0 2
# 7 b 33 mango 1 2
# 8 b 33 orange 0 2
#
# [[3]]
# name age fruit count tag
# 1 c 58 apple 1 1
# 2 c 58 cherry 1 1
# 3 c 58 mango 0 1
# 4 c 58 orange 0 1
# 5 c 58 apple 0 2
# 6 c 58 cherry 0 2
# 7 c 58 mango 0 2
# 8 c 58 orange 0 2
数据
vector1 <-
data.frame(
"name" = "a",
"age" = 10,
"fruit" = c("orange", "cherry", "apple"),
"count" = c(1, 1, 1),
"tag" = c(1, 1, 2),
stringsAsFactors = FALSE
)
vector2 <-
data.frame(
"name" = "b",
"age" = 33,
"fruit" = c("apple", "mango"),
"count" = c(1, 1),
"tag" = c(2, 2),
stringsAsFactors = FALSE
)
vector3 <-
data.frame(
"name" = "c",
"age" = 58,
"fruit" = c("cherry", "apple"),
"count" = c(1, 1),
"tag" = c(1, 1),
stringsAsFactors = FALSE
)
fruit_list <- list(vector1, vector2, vector3)
default <- c("cherry", "orange", "apple", "mango")