我们可以从每个日期获取月份和年份，并使用aggregate

aggregate(ids~month_year,transform(df,month_year = format(as.Date(edate),"%b-%Y")),length)

#  month_year ids
#1   Aug-2018   1
#2   Jul-2018   4

使用lubridate和dplyr的另一种方式

library(dplyr)
library(lubridate)

df %>%
  mutate(edate = as.Date(edate),year = year(edate),quarter = quarter(edate)) %>%
         #Use month to get monthly data
         #month = month(edate)) %>%
   count(year,quarter)

数据

df <- structure(list(edate = structure(c(3L,2L,4L,1L),.Label = c("2018-07-22","2018-07-24","2018-07-28","2018-08-03"),class = "factor"),ids = 2:6),class = "data.frame",row.names = c(NA,-5L))

Base R解决方案：

mthly_summary <- aggregate(list(mthly_tot = df$ids),by = list(mon_year = as.Date(df$edate,"%m-%Y")),sum)


qtrly_summary <- aggregate(list(qtrly_tot = df$ids),by = list(qtr_year = paste(gsub("[-].*","",df$edate),quarters(df$edate),sep = " ")),sum)

Tidyverse解决方案：

df %>%

  mutate(qtr = quarters(edate),mon_year = as.Date(edate,"%m-%Y")) %>% 

  group_by(qtr) %>%

  mutate(qtrly_to = sum(ids)) %>% 

  ungroup() %>% 

  group_by(mon_year) %>% 

  mutate(mthly_summary = sum(ids)) %>% 

  ungroup() %>%

  select(-c(ids,edate)) %>%

  distinct()

数据（感谢@Ronak Shah）：

df <- data.frame(edate = as.Date(c("2018-07-22","%Y-%m-%d"),ids = 2:5)