r - 如何柱状图星期几，并有字符串标签

Question

我有一个日期数据框（日期对象）；见底部。我正在尝试将它们转换为星期几，然后绘制直方图，但理想情况下标签是“星期一”...“星期日”（不是数字）

我有两个明显的问题：

1. 将 Date 对象转换为 day-of-week很容易，但结果是字符串或数字，而不是对象。
2. 当我得到直方图时，箱和标签是错误的（见下文）。

如果我使用weekdays(dat)，则输出是字符串 ("Monday"...) ，不能在hist().

或者，如果我转换为数字数据，如何获取字符串标签hist()？

> dotw <- with( month.day.year(dat[,1]), day.of.week(month,day,year) )
> hist(xxx,labels=c('M','Tu','W','Th','F','Sa','Su'),col='black') # WTF?!
> hist(dotw,xlab=list('M','Tu','W','Th','F','Sa','Su'))

不能按预期的方式工作。0.5 宽的箱子是怎么回事？还有，如何防止 Sunday->0 和 Monday->1 之间没有差距？理想情况下，列之间没有间隙。

我的数据看起来像：

> dat
  [1] "2010-04-02" "2010-04-06" "2010-04-09" "2010-04-10" "2010-04-14" "2010-04-15" "2010-04-19"
  [8] "2010-04-21" "2010-04-22" "2010-04-23" "2010-04-26" "2010-04-28" "2010-04-29" "2010-04-30"
 ...

> str(dat)
 Date[1:146], format: "2010-04-02" "2010-04-06" "2010-04-09" "2010-04-10" "2010-04-14" "2010-04-15" ...

> str(weekdays(dat))
 chr [1:146] "Friday" "Tuesday" "Friday" "Saturday" "Wednesday" "Thursday" "Monday" ...
> hist(weekdays(dat))
Error in hist.default(weekdays(dat)) : 'x' must be numeric

Answer 1

dat <- as.Date( c("2010-04-02", "2010-04-06", "2010-04-09", "2010-04-10", "2010-04-14", 
       "2010-04-15", "2010-04-19",   "2010-04-21", "2010-04-22", "2010-04-23","2010-04-24", 
        "2010-04-25", "2010-04-26", "2010-04-28", "2010-04-29", "2010-04-30"))
 dwka <- format(dat , "%a")
 dwka
# [1] "Fri" "Tue" "Fri" "Sat" "Wed" "Thu" "Mon"
#  [8] "Wed" "Thu" "Fri" "Sat" "Sun" "Mon" "Wed"
# [15] "Thu" "Fri"
dwkn <- as.numeric( format(dat , "%w") ) # numeric version
hist( dwkn , breaks= -.5+0:7, labels= unique(dwka[order(dwkn)]))

Answer 2

我怀疑你想要一个barplot而不是直方图。你可以table用来计算天数。

barplot(table(weekdays(dat)))

请注意，默认情况下，日期将按字母顺序排序，因此要更自然地排序，您必须在因子调用中重新排序级别：

barplot(table(factor(weekdays(dat),levels=c("Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"))))

Answer 3

将您转换weekdays(dat)为一个因子（分类变量的数据类型），并为直方图取消分类（将转换为整数）。因子类上有一些操作，可以轻松创建自定义 x 轴。

## days of the week
days <- c('Sun','Mon','Tues','Wed','Thurs','Fri','Sat')

## sample with replacement to generate data for this example
samples <- sample(days,100,replace=TRUE)

## convert to factor
## specify levels to specify the order
samples <- factor(samples,levels=days)

hist(unclass(samples),xaxt="n")
axis(1,at=1:nlevels(samples),lab=levels(samples))
box()