Question

我是R的初学者，也是R中data.table功能的忠实粉丝这可能是我在论坛上关于R的第一个问题，并对任何格式错误的代码/文本道歉。

我碰巧使用了大量分类变量的数据集我试图创建一个模拟数据集来解释这个问题：

structure(list(ID = 1234:1237, 
AgeC = c("25-30", "31+", "25-30", "20-24"), 
GenderC = c("female", "male", "female", "female"), 
doyoubuyappleseveryday = c(NA, 1L, NA, NA), 
doyoubuyapplesonceinaweek = c(1L, NA, NA, NA), 
doyoubuyapplesonceinamonth = c(NA, NA, NA, NA), 
doyoubuypearseveryday = c(NA, NA, NA, NA), 
doyoubuypearsonceinaweek = c(NA, NA, NA, 1L), 
doyoubuypearssonceinamonth = c(NA, NA, NA, NA), 
doyoueatappleseveryday = c(NA, NA, NA, NA), 
doyoueatapplesonceinaweek = c(1L, NA, 1L, NA), 
doyoueatapplesonceinamonth = c(NA, NA, NA, NA), 
doyoueatpearseveryday = c(NA, 1L, NA, NA), 
doyoueatpearsonceinaweek = c(NA, NA, NA, 1L), 
doyoueatpearsonceinamonth = c(1L, NA, NA, NA)), 
.Names = c("ID", "AgeC", "GenderC", "doyoubuyappleseveryday", 
    "doyoubuyapplesonceinaweek", "doyoubuyapplesonceinamonth", "doyoubuypearseveryday", 
    "doyoubuypearsonceinaweek", "doyoubuypearssonceinamonth", "doyoueatappleseveryday", 
    "doyoueatapplesonceinaweek", "doyoueatapplesonceinamonth", "doyoueatpearseveryday", 
    "doyoueatpearsonceinaweek", "doyoueatpearsonceinamonth"), 
    row.names = c(NA, -4L), class = c("data.table", "data.frame"), .internal.selfref = <pointer: 0x0000000000380788>)

我试图用选定的列重塑数据集..使用melt（）/ dcast（）

这个想法是，通过一系列的melt（）/ dcast（）操作，我可以通过创建可以获取分类值的适当新变量来将变量减少到6.

在第一组中，我选择使用“熔化（）/ dcast”函数消除“买”作为匹配字符串的列，如下所示：

id.cols.buy.m <-names(repro.dt)[-grep("buy",names(repro.dt))]
repro.buy.m<-data.table(melt(repro.dt, id.vars = id.cols.buy.m, 
                        measure.vars = grep("buy",names(repro.dt),value=T ),
                        na.rm=T,variable.name = "buy.fruit", 
                        value.name = "buy.freq"))

结果数据集会丢弃ID = 1236的整行，因为此ID的“购买”变量都是“NA”。这对我来说是一个数据丢失，因为在下一个重构操作序列中，“eat”作为匹配字符串，我打算使用之前的reshape / dcast中的清理版数据集。

如下：

#create a new variable "fruit"
repro.buy.m$fruit = apply(repro.buy.m, 1, function(u){
bool = sapply(fruit[,1], function(x) grepl(x,u[['buy.fruit']]))
if(any(bool)) fruit[bool] else NA
})


#create a fruit purchase freq column 

repro.buy.m$freq.pur= apply(repro.buy.m, 1, function(u){
bool = sapply(freq.levels[,1], function(x) grepl(x, u[['buy.fruit']]))
if(!is.na(u[['buy.freq']])) freq.levels[bool] else NA
})

#now drop the redundant colums that have "buy"  as match string--

id.cols.buy.m <- colnames(repro.buy.m)[-grep("buy",names(repro.buy.m))]
f <- as.formula(paste(paste(id.cols.buy.m, collapse = " + "), "~ buy.fruit"))

repro.buy.c<-data.table(dcast(data = repro.buy.m, f, 
                              value.var   ="buy.freq",
                              function(x)   length(unique(x))))

repro.buy.c<-repro.buy.c[, which(grepl("buy", colnames(repro.buy.c))):=NULL]

通过上述步骤，我所有的“购买”变量 - 其中6个现在减少到2个 然而，我丢失了一个不买水果但吃水果的身份证 如果我不使用na.rm = T，则保留行，但是当我使用“eat”作为匹配字符串重塑变量时，它会引发另一个重复行的问题

我的最终目标是将分类变量与名为“fruit”的单个列以及相关的列“freq.pur”和“freq.eat”合并为NA，并按ID进行分组。

类似的东西（这是一个ID）：

structure(list(fruit = c("apples", "pears"), freq.pur = c("onceinaweek",NA),
freq.et = c("onceinaweek", "onceinamonth")), 
.Names = c("fruit", "freq.pur", "freq.et"), 
row.names = c(NA, -2L), class = c("data.table", "data.frame"),
.internal.selfref = <pointer: 0x0000000000380788>, sorted = "fruit")

使用此示例数据集，我会在

上请求任何帮助

如何在部分和期间删除所选列的NA值顺序重塑数据 - 以避免重塑后的重复行
如何合并同一数据表中不同长度的列，这些数据表由单个分类变量连接，具有唯一值（在本例中为“fruit”），按ID分组 - 再次删除无意义的行

最好的问候

编辑：04/21：10：15IST（预期输出的模拟数据表）

structure(list(ID = c(1234L, 1234L, 1235L, 1235L, 1236L, 1237L), 
AgeC = c("25-30", "25-30", "31+", "31+", "25-30", "20-24"), 
GenderC = c("female", "female", "male", "male", "female", 
"female"), freq.pur = c(NA, "onceinaweek", "everyday", "everyday", 
NA, "onceinaweek"), freq.et = c("onceinamonth", "onceinaweek", 
"everyday", "everyday", "onceinaweek", "onceinaweek"), fruit = c("pears", 
"apples", "apples", "pears", "apples", "pears")), 
.Names = c("ID","AgeC", "GenderC", "freq.pur", "freq.et", "fruit"),
row.names = c(NA,-6L), class = c("data.table", "data.frame"),

.internal.selfref =）

编辑：04/25

我能够通过对中间结构和几个辅助变量的一些“独特”调用来解决这个问题，并且我能够测试所有用例。我的最终数据表如下：

dput(repro.buy.eat.final)

structure(list(ID = c(1210L, 1210L, 1234L, 1234L, 1234L, 1235L, 
1235L, 1237L, 1237L, 1238L, 1238L, 1239L), 
AgeC = c("25-30", "25-30", "25-30", "25-30", "25-30", "31+", 
"31+", "20-24", "20-24", "25-30", "25-30", "25-30"), 
GenderC = c("female", "female", "female", "female", "female", "male", 
"male", "female", "female", "male", "male", "male"), 
fruit = c("apples", "apples", "apples", "pears", "apples", "apples",
"pears", "pears", "pears", "apples", "pears", "pears"), 
freq.et = c("everyday", NA, NA, "onceinamonth", "onceinaweek", NA,
"everyday", "onceinaweek", NA, "onceinamonth", NA, NA), 
freq.pur = c(NA, "onceinaweek", "onceinaweek", NA, NA, "everyday", NA, 
NA,  "onceinaweek", NA, "everyday", "onceinaweek")), 
row.names = c(NA, -12L), 
class = c("data.table", "data.frame"),
.internal.selfref = <pointer: 0x0000000002830788>, 
.Names = c("ID", "AgeC", "GenderC", "fruit", "freq.et", "freq.pur"))

在结果集中，我想合并类似的水果与类似的“买”和“吃”频率。我在这里找到了一些相关答案： R: Merge of rows in same data table, concatenating certain columns，但是，我不知道如何应用条件来匹配频率，尽管我可以按ID和水果进行分组。我在这里寻求帮助。我可以在这里分享代码片段..如果这篇文章没有变得那么久。我还没有抓住格式化样式来在表格视图中附加数据表。

测试用例，我用过：

不买水果，但吃水果 -
没有吃水果，但买了水果
购买和吃同样的水果同样的频率
购买和吃同样的水果不同的频率
购买和吃diff水果相同的频率
买和吃diff fruits diff freq

最好的问候

Answer 1

这是一种方式。假设您的data.table被称为dt：

第1步：

以我们稍后可以拆分的方式更改列名：

setnames(dt, gsub("doyou(buy|eat)(apples|pears)", "\\1_\\2_", names(dt)))

\\1和\\2 在[{1}}的第一个参数中捕获在paranthesis ()中提供的值。基本上我已将gsub添加到我想要的位置，以便现在名称为：

_

现在我们可以分开names(dt) # [1] "ID" "AgeC" "GenderC" # [4] "buy_apples_everyday" "buy_apples_onceinaweek" "buy_apples_onceinamonth" # [7] "buy_pears_everyday" "buy_pears_onceinaweek" "buy_pears_sonceinamonth" # [10] "eat_apples_everyday" "eat_apples_onceinaweek" "eat_apples_onceinamonth" # [13] "eat_pears_everyday" "eat_pears_onceinaweek" "eat_pears_onceinamonth"。我建议使用分隔符以便于阅读。

第2步：

融化你的数据。表：

_

阅读警告并尝试理解。它会发出警告，因为您的某些列都是dt.m = melt(dt, id=names(dt)[1:3])（它们被加载为逻辑类型，而不是整数类型）。这很好，因为最终结果是整数列。所以你可以忽略警告。

第3步：

拆分 NA上的变量列，并创建3个单独的列：

_

dt.m[, c("buy_eat", "fruit", "freq") := tstrsplit(variable, "_", fixed=TRUE)] dt.m[, variable := NULL]在:=和列表上获取LHS中相应值的字符向量（对于列名称）。 RHS已经返回列表。通过执行以下操作分别检查tstrsplit的输出：tstrsplit(...)以了解它正在做什么。

tstrsplit(dt.m$variable, "_", fixed=TRUE)删除了该列。我们不再需要:= NULL了。所以我们删除它。当LHS是单个值时，为方便起见，我们不一定必须提供variable，即""和"variable"在此处的含义相同。

第4步：

variable value的位置，将NA替换为value，将0替换为freq：

NA

第5步：

对于每个dt.m[is.na(value), c("value", "freq") := list(0, NA)]，提取与最大ID, buy_eat, fruit对应的行。

value

ans = dt.m[, .SD[which.max(value)], by=.(ID, buy_eat, fruit)] # ID buy_eat fruit AgeC GenderC value freq # 1: 1234 buy apples 25-30 female 1 onceinaweek # 2: 1235 buy apples 31+ male 1 everyday # 3: 1236 buy apples 25-30 female 0 NA # 4: 1237 buy apples 20-24 female 0 NA # 5: 1234 buy pears 25-30 female 0 NA # 6: 1235 buy pears 31+ male 0 NA # 7: 1236 buy pears 25-30 female 0 NA # 8: 1237 buy pears 20-24 female 1 onceinaweek # 9: 1234 eat apples 25-30 female 1 onceinaweek # 10: 1235 eat apples 31+ male 0 NA # 11: 1236 eat apples 25-30 female 1 onceinaweek # 12: 1237 eat apples 20-24 female 0 NA # 13: 1234 eat pears 25-30 female 1 onceinamonth # 14: 1235 eat pears 31+ male 1 everyday # 15: 1236 eat pears 25-30 female 0 NA # 16: 1237 eat pears 20-24 female 1 onceinaweek返回which.max(<all_NA_values>)（长度为0的整数），这是不合需要的。这就是我们在上一步中将integer(0)替换为value 0的原因。

最后一步：

NA它。

dcast

我认为这是您正在寻找的结果。如果没有，我认为这应该让你对如何处理问题有所了解。我会把剩下的修修补补给你。

使用na.rm = T在选定列上进行熔合后，会完全删除数据行，尽管其余列具有有效数据

1 个答案: