在Stata中，我可以使用codebookout命令创建一个Excel工作簿，该工作簿可以保存现有数据集中所有变量的名称，标签和存储类型及其对应的值和值标签。
我想在R中找到一个等效的函数。到目前为止，我遇到了memisc库，它有一个名为codebook的函数，但它与Stata中的函数不同。
例如，在Stata中，码本的输出看起来像这样......（见下文 - 这就是我想要的）
Variable Name Variable Label Answer Label Answer Code Variable Type
hhid hhid Open ended String
inter_month inter_month Open ended long
year year Open ended long
org_unit org_unit long
Balaka 1
Blantyre 2
Chikwawa 3
Chiradzulu 4
即。评估数据框中的每一列以产生5个不同列的值：
这是我的尝试：
CreateCodebook <- function(dF){
numbercols <- length(colnames(dF))
table <- data.frame()
for (i in 1:length(colnames(dF))){
AnswerCode <- if (sapply(dF, is.factor)[i]) 1:nrow(unique(dF[i])) else ""
AnswerLabel <- if (sapply(dF, is.factor)[i]) unique(dF[order(dF[i]),][i]) else "Open ended"
VariableName <- if (length(AnswerCode) - 1 > 1) c(colnames(dF)[i],
rep("",length(AnswerCode) - 1)) else colnames(dF)[i]
VariableLabel <- if (length(AnswerCode) - 1 > 1) c(colnames(dF)[i],
rep("",length(AnswerCode) - 1)) else colnames(dF)[i]
VariableType <- if (length(AnswerCode) - 1 > 1) c(sapply(dF, class)[i],
rep("",length(AnswerCode) - 1)) else sapply(dF, class)[i]
df = data.frame(VariableName, VariableLabel, AnswerLabel, AnswerCode, VariableType)
names(df) <- c("Variable Name", "Variable Label", "Variable Type", "Answer Code", "Answer Label")
table <- rbind(table, df)
}
return(table)
}
不幸的是，我收到以下警告信息：
Warning messages:
1: In `[<-.factor`(`*tmp*`, ri, value = 1:3) :
invalid factor level, NA generated
2: In `[<-.factor`(`*tmp*`, ri, value = 1:2) :
invalid factor level, NA generated
我生成的输出导致答案代码标签搞砸了：
Variable Name Variable Label Variable Type Answer Code Answer Label
hhid hhid hhid Open ended character
month month month Open ended integer
year year year Open ended integer
org_unit org_unit org_unit Open ended character
v000 v000 v000 Open ended character
v001 v001 v001 Open ended integer
v002 v002 v002 Open ended integer
v003 v003 v003 Open ended integer
v005 v005 v005 Open ended integer
v006 v006 v006 Open ended integer
v007 v007 v007 Open ended integer
v021 v021 v021 Open ended numeric
2285 v024 v024 central <NA> factor
1 north <NA>
7119 south <NA>
11 v025 v025 rural <NA> factor
1048 v025 v025 urban <NA> factor
district_name district_name district_name Open ended character
coords_x1 coords_x1 coords_x1 Open ended numeric
coords_x2 coords_x2 coords_x2 Open ended numeric
itn_color itn_color itn_color Open ended numeric
piped piped piped Open ended numeric
sanit sanit sanit Open ended numeric
sanit_cd sanit_cd sanit_cd Open ended numeric
water water water Open ended numeric
答案 0 :(得分：2)
为了自己的娱乐，我决定采取行动。我使用了内置的Titanic数据集。但是我对你的一个定义有一个问题：你说“如果没有唯一的值，它就被认为是开放式的”。但长度> 0的每个变量都有一些独特的值：你的意思是“如果每个值都是唯一的”吗？即使这个定义也不一定按预期工作：在Titanic数据集中，响应是整数，并且在32个总值中恰好只有22个唯一值。我不认为人们真的希望枚举它，所以我测试了factor的类型（但如果你真的想要，可以用下面的length(u)==length(x)代替）。
## utility function: pad vector with blanks to specified length
pad <- function(x,n,p="") {
return(c(x,rep(p,n-length(x))))
}
## process a single column
proc_col <- function(x,nm) {
u <- unique(x)
## if (length(u)==length(x)) {
if (!is.factor(x)) {
n <- 1
u <- "open ended"
cc <- ""
} else {
cc <- as.numeric(u)
n <- length(u)
}
dd <- data.frame(`Variable Name`=pad(nm,n),
`Variable Label`=pad(nm,n),
`Answer Label`=u,
`Answer Code`=cc,
`Variable Type`=pad(class(x),n),
stringsAsFactors=FALSE)
return(dd)
}
## process all columns
proc_df <- function(x) {
L <- Map(proc_col,x,names(x))
dd <- do.call(rbind,L)
rownames(dd) <- NULL
return(dd)
}
示例：
xx <- as.data.frame.table(Titanic)
proc_df(xx)
## Variable.Name Variable.Label Answer.Label Answer.Code Variable.Type
## 1 Class Class 1st 1 factor
## 2 2nd 2
## 3 3rd 3
## 4 Crew 4
## 5 Sex Sex Male 1 factor
## 6 Female 2
## 7 Age Age Child 1 factor
## 8 Adult 2
## 9 Survived Survived No 1 factor
## 10 Yes 2
## 11 Freq Freq open ended numeric
我没有在代码值列表等之前留空格，但你可以自己进行调整......
答案 1 :(得分：0)
以下是解决问题的方法：
CreateCodebook <- function(dF){
numbercols <- length(colnames(dF))
table <- data.frame()
for (i in 1:length(colnames(dF))){
AnswerCode <- if (sapply(dF, is.factor)[i]) 1:nrow(unique(dF[i])) else ""
AnswerLabel <- if (sapply(dF, is.factor)[i]) unique(dF[order(dF[i]),][i]) else "Open ended"
VariableName <- if (length(AnswerCode) > 1) c(colnames(dF)[i],
rep("",length(AnswerCode) - 1)) else colnames(dF)[i]
VariableLabel <- if (length(AnswerCode) > 1) c(colnames(dF)[i],
rep("",length(AnswerCode) - 1)) else colnames(dF)[i]
VariableType <- if (length(AnswerCode) > 1) c(sapply(dF, class)[i],
rep("",length(AnswerCode) - 1)) else sapply(dF, class)[i]
df = data.frame(VariableName, VariableLabel, AnswerLabel, AnswerCode, VariableType, stringsAsFactors = FALSE)
names(df) <- c("Variable Name", "Variable Label", "Variable Type", "Answer Code", "Answer Label")
table <- rbind(table, df)
}
rownames(table) <- 1:nrow(table)
return(table)
}
输出：
Variable Name Variable Label Variable Type Answer Code Answer Label
1 brid brid Open ended character
2 month month Open ended integer
3 year year Open ended integer
4 org_unit org_unit Open ended character
5 v000 v000 Open ended character
6 v001 v001 Open ended integer
7 v002 v002 Open ended integer
8 v003 v003 Open ended integer
9 v005 v005 Open ended integer
10 v006 v006 Open ended integer
11 v007 v007 Open ended integer
12 v021 v021 Open ended numeric
13 v024 v024 central 1 factor
14 north 2
15 south 3
16 v025 v025 rural 1 factor
17 urban 2
18 bidx bidx Open ended integer
19 district_name district_name Open ended character
20 coords_x1 coords_x1 Open ended numeric
21 coords_x2 coords_x2 Open ended numeric
22 anc4 anc4 Open ended numeric
23 antimal_48 antimal_48 Open ended numeric
24 carep carep Open ended numeric
25 csec csec Open ended numeric
26 dptv dptv Open ended numeric
27 ebreast ebreast Open ended numeric
28 fans_48 fans_48 Open ended numeric
29 ideliv ideliv Open ended numeric
30 iptp iptp Open ended numeric
31 iron90 iron90 Open ended numeric
32 measlesv measlesv Open ended numeric
33 ors ors Open ended numeric
34 ort ort Open ended numeric
35 pncwm pncwm Open ended numeric
36 sstools sstools Open ended numeric
37 tt tt Open ended numeric
38 vita vita Open ended numeric