Question

假设我有以下数据：

set.seed(123)
timeseq <- as.Date(Sys.time() + cumsum(runif(1000)*86400))
data <- sample(c("A","B","C"), 1000,replace = TRUE)
data.frame(timeseq,data)

有谁知道如何按周查找A，B，C的计数？

Answer 1

您可以使用lubridate来解决此问题。因此将其与dplyr：

一起使用

library(lubridate)

data.frame(timeseq,data) %>%
  mutate(week=floor_date(timeseq, "week")) %>%
  group_by(data, week) %>%
  tally

其输出如下：

Source: local data frame [216 x 3]
Groups: data

   data       week n
1     A 2015-05-24 4
2     A 2015-05-31 3
3     A 2015-06-07 3
4     A 2015-06-14 5
5     A 2015-06-21 6
6     A 2015-06-28 3
7     A 2015-07-05 8
8     A 2015-07-12 3
9     A 2015-07-19 4
10    A 2015-07-26 6

Answer 2

如果您想根据周数回答，请尝试使用

set.seed(123)
timeseq <- as.Date(Sys.time() + cumsum(runif(1000)*86400))
data <- sample(c("A","B","C"), 1000,replace = TRUE)
data.frame(timeseq,data)

mydf <- as.Date(timeseq, format="%d-%m-%Y")
weeknum <- as.numeric( format(mydf+3, "%U"))
weeknum

new_data <- data.frame(timeseq,weeknum,data)

table(new_data$weeknum,new_data$data)

R中每周因子频率的总和

2 个答案: