分组按日期数据帧：解决失踪时间段错误
问题描述：
我已经确定，如果不是自己创建的，困难的错误在一些不错的代码解决从这里StackOverflow上慷慨申请人收到几个星期前，我今天可以使用一些新的援助。分组按日期数据帧：解决失踪时间段错误
样本数据（下面称为对象eh）：
ID 2013-03-20 2013-04-09 2013-04-11 2013-04-17 2013-04-25 2013-05-15 2013-05-24 2013-05-25 2013-05-26
5167f 0 0 0 0 0 0 0 0 0
1214m 0 0 0 0 0 0 0 0 0
1844f 0 0 0 0 0 0 0 0 0
2113m 0 0 0 0 0 0 0 0 0
2254m 0 0 0 0 0 0 0 0 0
2721f 0 0 0 0 0 0 0 0 0
3121f 0 0 0 0 0 0 0 0 0
3486f 0 0 0 0 0 0 0 0 0
3540f 0 0 0 0 0 0 0 0 0
4175m 0 0 0 0 0 0 0 0 0
我需要能够组0s和1s由其中它们各自的柱日期落在时间周期（例如，每隔1，2，3 ，或4周）。每当1在特定日期范围（Period）内至少下降一次，则在该Period（0，否则）中针对该ID总结1。
我开始用1个星期的例行总结作为一个例子。我的主要问题是，所产生的最终输出的时间序列"2013-03-20"到"2015-12-31"中缺少一些Periods总数可能1周。在该例子中的输出，其中，所述行是唯一IDs和列是用于唯一Periods，如何Periods 2，5，7，和9被丢失
说明：
1 3 4 6 8 10 11 12 13 14
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
这里是充分例程分组原始数据帧（请参阅上面共享的示例数据）：
#Convert to data table from original data frame, eh
dt <- as.data.table(eh)
#One week summarized encounter histories
dt_merge <- data_frame(
# Create a column showing the beginning date
Date1 = seq(from = ymd("2013-03-20"), to = ymd("2015-12-31"), by = "1 week")) %>%
# Create a column showing the end date of each period
mutate(Date2 = lead(Date1)) %>%
# Adjust Date1
mutate(Date1 = if_else(Date1 == ymd("2013-03-20"), Date1, Date1 + 1)) %>%
# Remove the last row
drop_na(Date2) %>%
# Create date list
mutate(Dates = map2(Date1, Date2, function(x, y){ seq(x, y, by = "day") })) %>%
unnest() %>%
# Create Group ID
mutate(RunID = group_indices_(., dots. = c("Date1", "Date2"))) %>%
# Create Period ID
mutate(Period = paste0(RunID)) %>%
# Add a column showing Month
mutate(Month = month(Dates)) %>%
# Add a column showing Year
mutate(Year = year(Dates)) %>%
# Add a column showing season
mutate(Season = case_when(
Month %in% 3:5 ~ "Spring",
Month %in% 6:8 ~ "Summer",
Month %in% 9:11 ~ "Fall",
Month %in% c(12, 1, 2) ~ "Winter",
TRUE ~ NA_character_
)) %>%
# Combine Season and Year
mutate(SeasonYear = paste0(Season, Year)) %>%
select(-Date1, -Date2, -RunID)
dt2 <- dt %>%
# Reshape the data frame
gather(Date, Value, -ID) %>%
# Convert Date to date class
mutate(Date = ymd(Date)) %>%
# Join dt_merge
left_join(dt_merge, by = c("Date" = "Dates"))
one.week <- dt2 %>%
group_by(ID, Period) %>%
summarise(Value = max(Value)) %>%
spread(Period, Value)
#Finished product
one.week <- as.data.frame(one.week)
#Missing weeks 2, 5, 7, and 9...
one.week
有人可以帮助我了解我哪里出错了吗？提前致谢！
〜AD
答
发生这种情况，因为这些周从eh数据丢失。例如，如果你看一下，使上涨2周日期：
dt_merge %>%
filter(Period == 2)
#> # A tibble: 7 x 6
#> Dates Period Month Year Season SeasonYear
#> <date> <chr> <dbl> <dbl> <chr> <chr>
#> 1 2013-03-28 2 3 2013 Spring Spring2013
#> 2 2013-03-29 2 3 2013 Spring Spring2013
#> 3 2013-03-30 2 3 2013 Spring Spring2013
#> 4 2013-03-31 2 3 2013 Spring Spring2013
#> 5 2013-04-01 2 4 2013 Spring Spring2013
#> 6 2013-04-02 2 4 2013 Spring Spring2013
#> 7 2013-04-03 2 4 2013 Spring Spring2013
你可以看到，没有这些日期都在eh列，其中跳过从2013年3月20日至2013-04- 09。因为你创造dt2，只有在eh日期（因此周）时使用left_join被保留。
这可以通过使用tidyr包中的complete()来更正ID和Date的缺失组合。
dt2 <- dt %>%
# Reshape the data frame
gather(Date, Value, -ID) %>%
# Convert Date to date class
mutate(Date = ymd(Date)) %>%
# Create missing ID/Date combinations
complete(ID, Date = dt_merge$Dates) %>%
# Join dt_merge
left_join(dt_merge, by = c("Date" = "Dates"))
one.week <- dt2 %>%
mutate(Period = as.numeric(Period)) %>%
group_by(ID, Period) %>%
summarise(Value = max(Value, na.rm = TRUE)) %>%
spread(Period, Value)
one.week
#> # A tibble: 10 x 146
#> # Groups: ID [10]
#> ID `1` `2` `3` `4` `5` `6` `7` `8` `9` `10` `11`
#> * <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 1214m 0 -Inf 0 0 -Inf 0 -Inf 0 -Inf 0 -Inf
#> 2 1844f 0 -Inf 0 0 -Inf 0 -Inf 0 -Inf 0 -Inf
#> 3 2113m 0 -Inf 0 0 -Inf 0 -Inf 0 -Inf 0 -Inf
#> 4 2254m 0 -Inf 0 0 -Inf 0 -Inf 0 -Inf 0 -Inf
#> 5 2721f 0 -Inf 0 0 -Inf 0 -Inf 0 -Inf 0 -Inf
#> 6 3121f 0 -Inf 0 0 -Inf 0 -Inf 0 -Inf 0 -Inf
#> 7 3486f 0 -Inf 0 0 -Inf 0 -Inf 0 -Inf 0 -Inf
#> 8 3540f 0 -Inf 0 0 -Inf 0 -Inf 0 -Inf 0 -Inf
#> 9 4175m 0 -Inf 0 0 -Inf 0 -Inf 0 -Inf 0 -Inf
#> 10 5167f 0 -Inf 0 0 -Inf 0 -Inf 0 -Inf 0 -Inf
#> # ... with 134 more variables: `12` <dbl>, `13` <dbl>, `14` <dbl>,
#> # `15` <dbl>, `16` <dbl>, `17` <dbl>, `18` <dbl>, `19` <dbl>,
#> # `20` <dbl>, `21` <dbl>, `22` <dbl>, `23` <dbl>, `24` <dbl>,
#> # `25` <dbl>, `26` <dbl>, `27` <dbl>, `28` <dbl>, `29` <dbl>,
#> # `30` <dbl>, `31` <dbl>, `32` <dbl>, `33` <dbl>, `34` <dbl>,
#> # `35` <dbl>, `36` <dbl>, `37` <dbl>, `38` <dbl>, `39` <dbl>,
#> # `40` <dbl>, `41` <dbl>, `42` <dbl>, `43` <dbl>, `44` <dbl>,
#> # `45` <dbl>, `46` <dbl>, `47` <dbl>, `48` <dbl>, `49` <dbl>,
#> # `50` <dbl>, `51` <dbl>, `52` <dbl>, `53` <dbl>, `54` <dbl>,
#> # `55` <dbl>, `56` <dbl>, `57` <dbl>, `58` <dbl>, `59` <dbl>,
#> # `60` <dbl>, `61` <dbl>, `62` <dbl>, `63` <dbl>, `64` <dbl>,
#> # `65` <dbl>, `66` <dbl>, `67` <dbl>, `68` <dbl>, `69` <dbl>,
#> # `70` <dbl>, `71` <dbl>, `72` <dbl>, `73` <dbl>, `74` <dbl>,
#> # `75` <dbl>, `76` <dbl>, `77` <dbl>, `78` <dbl>, `79` <dbl>,
#> # `80` <dbl>, `81` <dbl>, `82` <dbl>, `83` <dbl>, `84` <dbl>,
#> # `85` <dbl>, `86` <dbl>, `87` <dbl>, `88` <dbl>, `89` <dbl>,
#> # `90` <dbl>, `91` <dbl>, `92` <dbl>, `93` <dbl>, `94` <dbl>,
#> # `95` <dbl>, `96` <dbl>, `97` <dbl>, `98` <dbl>, `99` <dbl>,
#> # `100` <dbl>, `101` <dbl>, `102` <dbl>, `103` <dbl>, `104` <dbl>,
#> # `105` <dbl>, `106` <dbl>, `107` <dbl>, `108` <dbl>, `109` <dbl>,
#> # `110` <dbl>, `111` <dbl>, ...
这里-Inf是，如果有在给定星期ID没有值返回。或者，也可以使用complete(ID, Date = dt_merge$Dates, fill = list(Value = 0))填充缺省值NA，以填充例如0。这将使任何不可观察的ID和日期组合的值变量0。
这是它。非常感谢你！ – Andrew