将时间表（宽格式）转换为 R 中的长格式并且 select 只有一些行

Question

我想为级别和日期的每个组合创建行，其中有 "X"。

示例：

dat <- data.frame(1:5, matrix( c("X", "-", "X", "-", "-", "X", "X", "-", "X", "-", "X", "-", "X", "-", "X", "-", "X", "-", "-", "-", "-", "-", "X", "X", "-"), 5, 5, byrow=T))
colnames(dat) <- c("lvl",  "Mon",  "Tues", "Wed",  "Thur", "Fri" )

Answer 1

如果您的时间表存储为矩阵或数据框，您可以使用 reshape2 包：

# generate a fake schedule
sched <- matrix(rbinom(25, 1, 1/2), 5,
  dimnames = list(1:5, c("Mo", "Tu", "We", "Th", "Fr")))

library(reshape2)
melt(sched)

#    Var1 Var2 value
# 1     1   Mo     0
# 2     2   Mo     0
# 3     3   Mo     1
# 4     4   Mo     0
# 5     5   Mo     1
# 6     1   Tu     0
# 7     2   Tu     1
# 8     3   Tu     1
# 9     4   Tu     0
# 10    5   Tu     1
# 11    1   We     1
# 12    2   We     1
# 13    3   We     1
# 14    4   We     0
# 15    5   We     0
# 16    1   Th     0
# 17    2   Th     1
# 18    3   Th     1
# 19    4   Th     0
# 20    5   Th     0
# 21    1   Fr     0
# 22    2   Fr     1
# 23    3   Fr     1
# 24    4   Fr     1
# 25    5   Fr     1

包tidyr中也有函数gather，但它只适用于数据帧，总体上更高级一些。

要仅获取 value 列中带有 1 的行，请执行：

sched <- melt(sched)
sched[sched$value == 1, ]

将时间表（宽格式）转换为 R 中的长格式并且 select 只有一些行

Convert a schedule (in wide form) to long form in R and select only some rows

r

data-manipulation