2016-10-25 122 views
-3

我有一些关于ID和开始日期组合的ID,日期和整数值的数据,每个ID有多个日期。查找数据表中的重叠

我想创建指示柱:

1)告诉我如果ID具有从整数,或4个独立的整数在12个月内的计数的总和> = 14。

这里有一个类似的问题,但我的类别是一个比较复杂: Create new column based on condition that exists within a rolling date

任何帮助,不胜感激!

下面是一些数据的dput:

structure(list(ID = c("90939293", "90963328", "90092983", 
"90032926", "90944838", "90092983", "90062392", "90224939", "90202398", 
"90926203", "90936043", "90329263", "90944838", "90232033", "90980903", 
"90924463", "90299292", "90933383", "90209349", "90092983", "90022988", 
"90022293", "90933383", "90092983", "90299240", "90963033", "90004923", 
"90292998", "90986096", "90980903", "90336692", "90933383", "90022988", 
"90069992", "90062392", "90209248", "90924463", "90092983", "90933383", 
"90022293", "90062392", "90004923", "90233269", "90329263", "90229202", 
"90309943", "90299292", "90036820", "90329263", "90232033", "90329263", 
"90336692", "90963033", "90224939", "90924463", "90069992", "90092983", 
"90934923", "90926203", "90222333", "90092983", "90299292", "90202398", 
"90004923", "90233269", "90926203", "90222333", "90224939", "90232033", 
"90933383", "90022293", "90022988", "90934923", "90069992", "90329263", 
"90209349", "90022293", "90309943", "90299240", "90022293", "90336692", 
"90020334", "90933383", "90290384", "90224939", "90980903", "90299240", 
"90299292", "90202398", "90022346"), Date = structure(c(15972, 
16009, 16010, 16010, 16007, 16010, 16006, 16010, 16007, 16008, 
15997, 16007, 16007, 16002, 16008, 16006, 16006, 16006, 16009, 
16010, 16006, 16006, 16006, 16010, 15995, 16008, 16008, 16010, 
16009, 16008, 16010, 16006, 16006, 16009, 16006, 16006, 16006, 
16010, 16006, 16006, 16006, 16008, 16009, 16007, 16010, 16007, 
16006, 16009, 16007, 16002, 16007, 16010, 16008, 16010, 16006, 
16009, 16010, 15936, 16008, 16008, 16010, 16006, 16007, 16008, 
16009, 16008, 16008, 16010, 16002, 16006, 16006, 16006, 15936, 
16009, 16007, 16009, 16006, 16007, 15995, 16006, 16010, 16006, 
16006, 16010, 16010, 16008, 15995, 16006, 16007, 16008), class = "Date"), 
    Integer = c(39, 2, 1, 1, 4, 1, 5, 1, 4, 3, 14, 4, 4, 9, 
    3, 5, 5, 5, 2, 1, 5, 5, 5, 1, 16, 3, 3, 1, 2, 3, 1, 5, 5, 
    2, 5, 5, 5, 1, 5, 5, 5, 3, 2, 4, 1, 4, 5, 2, 4, 9, 4, 1, 
    3, 1, 5, 2, 1, 75, 3, 3, 1, 5, 4, 3, 2, 3, 3, 1, 9, 5, 5, 
    5, 75, 2, 4, 2, 5, 4, 16, 5, 1, 5, 5, 1, 1, 3, 16, 5, 4, 
    3)), .Names = c("ID", "Date", "Integer" 
), row.names = c("200086", "200066", "200050", "200064", "200078", 
"200050.1", "200069", "200082", "200083", "200053", "200056", 
"200055", "200078.1", "200079", "200051", "200089", "200052", 
"200057", "200061", "200050.2", "200060", "200080", "200057.1", 
"200050.3", "200068", "200071", "200070", "200059", "200062", 
"200051.1", "200067", "200057.2", "200060.1", "200072", "200069.1", 
"200073", "200089.1", "200050.4", "200057.3", "200080.1", "200069.2", 
"200070.1", "200081", "200054", "200063", "200075", "200052.1", 
"200074", "200054.1", "200079.1", "200055.1", "200067.1", "200071.1", 
"200082.1", "200089.2", "200072.1", "200050.5", "200084", "200053.1", 
"200088", "200050.6", "200052.2", "200083.1", "200070.2", "200081.1", 
"200053.2", "200088.1", "200082.2", "200079.2", "200057.4", "200080.2", 
"200060.2", "200084.1", "200072.2", "200055.2", "200061.1", "200080.3", 
"200075.1", "200068.1", "200080.4", "200067.2", "200065", "200057.5", 
"200090", "200082.3", "200051.2", "200068.2", "200052.3", "200083.2", 
"200076"), class = "data.frame") 
+5

“有每个ID多个日期” - '任何(复制(DF $ X1))'你不同意,你的样本数据。你的ID(第一列,我假设它们在你的例子中只叫X1)是唯一的。或者你的意思是某些日期有多个ID?无论哪种方式,请制作一个**小**示例而不是100行。 – Spacedman

+0

这并不明确:“告诉我一个ID在12个月内是否有14个整数或4个独立整数的总和”。 “14个整数的和”是什么意思? 1 + 2 + 3 + 4 + 1 + 2 + 3 + 4 + 1 + 2 + 3 + 4 + 7 + 99是14个整数的和。你不是那个意思吗? – Spacedman

+0

我认为你可能在这里提出太多问题,所以不鼓励部分答案,所以除非有一个人解决你的所有问题,否则你将不会得到任何答案。建议你删除这个帖子,并创建几个 - 第一个将是如何找到哪些ID的总和的“整数”列值等于14. – Spacedman

回答

1

与dput为x:

library(data.table) 

setDT(x, key = "Date") 

# test 1 
x[, `:=` (
    test1 = sum(Integer) >= 14 
), by = ID] 

# test2 
y = x[, .(
    count12 = uniqueN(Integer) 
), by = .(start = Date, end = Date - 365)] 

# combine 
z = merge(x, y, by.x = "Date", by.y = "start") 
z[, end := NULL] 
z[, flag := test1 | count12 == 4] 
1

这里是你要问的一个刺。现在,查找ID大于14的整数之和的ID就像按ID分组一样容易,并检查每个ID的整数列的总和是否大于等于14,或者检查dplyr:df %>% group_by(ID) %>% mutate(conditional = sum(Integer) >= 14)。在12个月的时间内找到(至少?)4的ID显然更难。我的解决方案如下this答案在计算窗口计数。

只有一个警告:由于roll_sum通过滚动行数来工作,所以我使用的解决方案依赖于每个ID每天只有一行。在你的示例数据框中,实际上有多个相同ID日期的条目,但它们看起来是重复的,所以我删除了它们。如果它们不是,并且需要为sum(Integer) >= 14的条件计算重复值,则可以事先对它们进行总结(例如:df %>% group_by(ID, Date) %>% summarize(Integer = sum(Integer))),以便每个日期每个ID只有一个条目。

library(dplyr) 
library(tidyr) 
library(RcppRoll) 

df_tmp <- df 
df <- df_tmp %>% 
    group_by(ID, Date) %>% 
    filter(n() == 1) %>% # this line removes duplicate columns 
    ungroup() %>% 
    complete(ID, 
      Date=seq(from=min(Date)-365,to=max(Date), by=1), 
      fill=list(Integer=0)) %>% # we use complete to add in a row for all IDs for every single date since a year before the first obs. 
    arrange(ID, Date) %>% 
    group_by(ID) %>% 
    mutate(roll_count = roll_sum(x = Integer != 0, n = 365, fill=0, align="right"), # this calculates the rolling sum using n = 365 as a stand-in for 12 months 
     conditional = sum(Integer) >= 14 || roll_count >= 4) %>% 
    ungroup() %>% 
    right_join(df, by = c("ID","Date", "Integer")) # right_join with the original data to remove dummy dates 

希望这有助于!