我们首先可以从startdate获得小时，每小时获得count行数，type得到小时。 complete缺少小时数，并用0填充其计数，然后使用pivot_wider获取宽格式的数据。

library(dplyr)
library(tidyr)

df %>%
  mutate(hr = lubridate::hour(startdate)) %>%
  count(hr,type) %>%
  complete(type,hr = seq(0,max(hr)),fill = list(n = 0)) %>%
  pivot_wider(names_from = type,values_from = n)

# A tibble: 4 x 3
#     hr version1 version2
#  <int>    <dbl>    <dbl>
#1     0        0        0
#2     1        0        0
#3     2        2        1
#4     3        1        1

开始日期变量出了点问题。因此，我使用软件包lubridate

library(dplyr)
library(tidyr)

type = c('version1','version1','version2','version2')

startdate = lubridate::ymd_hms(c('2017-11-1T02:11:02.000','2018-3-25T02:13:02.000','2019-3-14T03:45:02.000','2017-3-14T02:55:02.000','2018-3-14T03:45:02.000'))

tibble(type = type,startdate = startdate) %>%
  count(type,hour = lubridate::hour(startdate)) %>%
  spread(type,n)

# A tibble: 2 x 3
   hour version1 version2
  <int>    <int>    <int>
1     2        2        1
2     3        1        1

Base R解决方案：

# Extract the hour and store it as a vector: 

df$hour <- gsub(".* ","",trunc(df$startdate,units = "hours"))

# Count the number of observations of each type in each hour: 

df$type_hour_cnt <- with(df,ave(paste(type,hour,sep = " - "),paste(type,FUN = seq_along))

# Reshape dataframe: 

df <- as.data.frame(as.matrix(xtabs(type_hour_cnt ~ hour + type,df,sparse = T)))

# Extract rownames and store them as "hour" vector and then delete row.names: 

df <- data.frame(cbind(hour = row.names(df),df),row.names = NULL)