用分号分隔符分割数据帧？

Question

我想在不使用第三方包的情况下使用 R 在数据框中拆分以分号分隔的文本。

我有以下数据。

> #To view the first 6 rows of the data
> head(bank1)
  age.job.marital.education.default.housing.loan.contact.month.day_of_week.duration.campaign.pdays.previous.poutcome.emp.var.rate.cons.price.idx.cons.conf.idx.euribor3m.nr.employed.y
1                                                                      56;housemaid;married;basic.4y;no;no;no;telephone;may;mon;261;1;999;0;nonexistent;1.1;93.994;-36.4;4.857;5191;no
2                                                               57;services;married;high.school;unknown;no;no;telephone;may;mon;149;1;999;0;nonexistent;1.1;93.994;-36.4;4.857;5191;no
3                                                                   37;services;married;high.school;no;yes;no;telephone;may;mon;226;1;999;0;nonexistent;1.1;93.994;-36.4;4.857;5191;no
4                                                                         40;admin.;married;basic.6y;no;no;no;telephone;may;mon;151;1;999;0;nonexistent;1.1;93.994;-36.4;4.857;5191;no
5                                                                   56;services;married;high.school;no;no;yes;telephone;may;mon;307;1;999;0;nonexistent;1.1;93.994;-36.4;4.857;5191;no
6                                                                  45;services;married;basic.9y;unknown;no;no;telephone;may;mon;198;1;999;0;nonexistent;1.1;93.994;-36.4;4.857;5191;no

请帮助我根据标题中的列名将数据分成不同的列。

提前致谢。

Answer 1

您可以将data.frame 列连接到character 字符串并再次运行read.table。但是请注意，列名的数量（28）不等于列的数量（21）。此外，列标题和观察的分隔符不同（看起来像标题的空格，观察的分号）。

请看下面的代码：

df <- structure(list(age.job.marital.education.default.housing.loan.contact.month.day_of_week.duration.campaign.pdays.previous.poutcome.emp.var.rate.cons.price.idx.cons.conf.idx.euribor3m.nr.employed.y = structure(c(4L, 
6L, 1L, 2L, 5L, 3L), .Label = c("37;services;married;high.school;no;yes;no;telephone;may;mon;226;1;999;0;nonexistent;1.1;93.994;-36.4;4.857;5191;no", 
"40;admin.;married;basic.6y;no;no;no;telephone;may;mon;151;1;999;0;nonexistent;1.1;93.994;-36.4;4.857;5191;no", 
"45;services;married;basic.9y;unknown;no;no;telephone;may;mon;198;1;999;0;nonexistent;1.1;93.994;-36.4;4.857;5191;no", 
"56;housemaid;married;basic.4y;no;no;no;telephone;may;mon;261;1;999;0;nonexistent;1.1;93.994;-36.4;4.857;5191;no", 
"56;services;married;high.school;no;no;yes;telephone;may;mon;307;1;999;0;nonexistent;1.1;93.994;-36.4;4.857;5191;no", 
"57;services;married;high.school;unknown;no;no;telephone;may;mon;149;1;999;0;nonexistent;1.1;93.994;-36.4;4.857;5191;no"
), class = "factor")), row.names = c(NA, -6L), class = "data.frame")

z <- paste(df[, 1], sep = "", collapse = "\n")
df2 <- read.table(text = z, header = FALSE, sep = ";")

nms2 <- unlist(strsplit(names(df), "\\."))[-(22:28)]
names(df2) <- nms2
str(df2)

输出：

'data.frame':   6 obs. of  21 variables:
 $ age        : int  56 57 37 40 56 45
 $ job        : Factor w/ 3 levels "admin.","housemaid",..: 2 3 3 1 3 3
 $ marital    : Factor w/ 1 level "married": 1 1 1 1 1 1
 $ education  : Factor w/ 4 levels "basic.4y","basic.6y",..: 1 4 4 2 4 3
 $ default    : Factor w/ 2 levels "no","unknown": 1 2 1 1 1 2
 $ housing    : Factor w/ 2 levels "no","yes": 1 1 2 1 1 1
 $ loan       : Factor w/ 2 levels "no","yes": 1 1 1 1 2 1
 $ contact    : Factor w/ 1 level "telephone": 1 1 1 1 1 1
 $ month      : Factor w/ 1 level "may": 1 1 1 1 1 1
 $ day_of_week: Factor w/ 1 level "mon": 1 1 1 1 1 1
 $ duration   : int  261 149 226 151 307 198
 $ campaign   : int  1 1 1 1 1 1
 $ pdays      : int  999 999 999 999 999 999
 $ previous   : int  0 0 0 0 0 0
 $ poutcome   : Factor w/ 1 level "nonexistent": 1 1 1 1 1 1
 $ emp        : num  1.1 1.1 1.1 1.1 1.1 1.1
 $ var        : num  94 94 94 94 94 ...
 $ rate       : num  -36.4 -36.4 -36.4 -36.4 -36.4 -36.4
 $ cons       : num  4.86 4.86 4.86 4.86 4.86 ...
 $ price      : int  5191 5191 5191 5191 5191 5191
 $ idx        : Factor w/ 1 level "no": 1 1 1 1 1 1