Question

我正在尝试使用多个条件合并2个数据框，并使用了合并命令但无法获得成功的输出。

#Data Frame df1#
ID<- c("A1", "A2","A3", "A4")
Location <- c("012A","234B","012A","238C" )
startdate <- as.Date(c("2014-11-01","2014-01-01","2015-10-01", "2015-01-01"))
enddate <- as.Date(c("2014-12-31","2014-08-31","2015-12-31","2015-12-31"))
df1<- data.frame(ID,Location, startdate, enddate)

#Data Frame df2#
ID<-c("A1", "A1", "A4")
N<- c(2,1,2)
Loss_Date <- as.Date(c("2014-11-15", "2015-12-25", "2015-11-30"))
Amt<-c("2200","1000", "500")
df2<- data.frame(ID, N, Loss_Date,Amt)

我想通过使用Location作为公共列合并这两个数据帧，df2中的Loss_Date位于df2中的（包括）Start_Date和End_Date之间。您可以看到df2中的第二个条目未被映射为日期不在df1范围内

#Required Output
ID<- c("A1", "A2","A3", "A4")
Location <- c("012A","234B","012A","238C" )
startdate <- as.Date(c("2014-11-01","2014-01-01","2015-10-01", "2015-01-01"))
enddate <- as.Date(c("2014-12-31","2014-08-31","2015-12-31","2015-12-31"))
N<-c(2,0,0,2)
Loss_Date <- c("2014-11-15", "NA", "NA", "2015-11-30")
Amt<-c("2200","0","0", "500")
Output<- data.frame(ID,Location, startdate, enddate,N, Loss_Date,Amt)

我使用Year和ID创建了一个公共ID，但是得到了错误的映射。尝试了各种使用合并和匹配的方法，但命令不起作用。我需要这个以超过170K的观察值运行。两个数据帧的长度都不相等。任何帮助将非常感激。

Answer 1

我在@VincentBoned的回答中添加了一些额外的代码。

# create 1st dataframe
ID<- c("A1", "A2","A3", "A4")
Location <- c("012A","234B","012A","238C" )
startdate <- as.Date(c("2014-11-01","2014-01-01","2015-10-01", "2015-01-01"))
enddate <- as.Date(c("2014-12-31","2014-08-31","2015-12-31","2015-12-31"))

df1<- data.frame(ID,Location, startdate, enddate, stringsAsFactors = F)


# create 2nd dataframe
ID<-c("A1", "A1", "A4")
N<- c(2,1,2)
Loss_Date <- as.Date(c("2014-11-15", "2015-12-25", "2015-11-30"))
Amt<-c("2200","1000", "500")

df2<- data.frame(ID, N, Loss_Date,Amt, stringsAsFactors = F)


library(dplyr)

full_join(df1, df2, by="ID") %>% 
  mutate(condition = (Loss_Date >= startdate & Loss_Date <= enddate)) %>%
  mutate(N = ifelse(condition & !is.na(condition), N, 0)) %>%
  mutate(Loss_Date = as.Date(ifelse(condition, Loss_Date, NA),origin="1970-01-01")) %>%
  mutate(Amt = ifelse(condition & !is.na(condition), Amt, 0)) %>%
  select(-condition) %>%
  group_by(ID) %>%                              # for each ID
  mutate(Nrows = n()) %>%                       # count how many rows they have in the final table
  ungroup() %>%
  filter(!(Nrows > 1 & is.na(Loss_Date))) %>%   # filter out rows with IDs that have more than 1 rows and those rows are not matched
  select(-Nrows)

#   ID Location  startdate    enddate N  Loss_Date  Amt 
# 1 A1     012A 2014-11-01 2014-12-31 2 2014-11-15 2200 
# 2 A2     234B 2014-01-01 2014-08-31 0       <NA>    0 
# 3 A3     012A 2015-10-01 2015-12-31 0       <NA>    0 
# 4 A4     238C 2015-01-01 2015-12-31 2 2015-11-30  500

如果您了解上述代码的工作原理（一步一步），您可以使用更紧凑的版本，返回相同的结果：

full_join(df1, df2, by="ID") %>% 
  mutate(condition = (Loss_Date >= startdate & Loss_Date <= enddate),
         N = ifelse(condition & !is.na(condition), N, 0),
         Loss_Date = as.Date(ifelse(condition, Loss_Date, NA),origin="1970-01-01"),
         Amt = ifelse(condition & !is.na(condition), Amt, 0)) %>%
  group_by(ID) %>%                             
  mutate(Nrows = n()) %>%                      
  filter(!(Nrows > 1 & is.na(Loss_Date))) %>%
  select(-c(condition, Nrows))

Answer 2

在 data.table （v1.9.7）的当前开发版本中，实现了非equi连接。我们可以这样做：

require(data.table) # v1.9.7+
setDT(df2)[df1, .(ID, Location, startdate, enddate, N, x.Loss_Date, Amt), 
                      on=.(ID, Loss_Date>=startdate, Loss_Date<=enddate)]
#    ID Location  startdate    enddate  N x.Loss_Date  Amt
# 1: A1     012A 2014-11-01 2014-12-31  2  2014-11-15 2200
# 2: A2     234B 2014-01-01 2014-08-31 NA        <NA>   NA
# 3: A3     012A 2015-10-01 2015-12-31 NA        <NA>   NA
# 4: A4     238C 2015-01-01 2015-12-31  2  2015-11-30  500

Answer 3

我使用包dplyr完成了合并，这非常快速且易于使用。

您应该将此stringsAsFactors=F

添加到数据框定义中

 df1<- data.frame(ID,Location, startdate, enddate, stringsAsFactors = F)
 df2<- data.frame(ID, N, Loss_Date,Amt, stringsAsFactors = F)

因此，您的角色输入不会更改为因素，并且它们不会给您带来不良后果

install.packages("dplyr")
library(dplyr)

output <- full_join(df1, df2, by="ID") %>% 
filter(Loss_Date >= startdate & Loss_Date <= enddate)

输出：

  ID Location  startdate    enddate N  Loss_Date  Amt
1 A1     012A 2014-11-01 2014-12-31 2 2014-11-15 2200
2 A4     238C 2015-01-01 2015-12-31 2 2015-11-30  500

同样，根据评论的指定，如果要保留与条件不匹配的行，则应使用其他函数：

output2 <- left_join(df1, df2, by="ID") %>% 
 mutate(condition = (Loss_Date >= startdate & Loss_Date <= enddate)) %>%
 mutate(N = ifelse(condition & !is.na(condition), N, 0)) %>%
 mutate(Loss_Date = as.Date(ifelse(condition, Loss_Date, NA),origin="1970-01-01")) %>%
 mutate(Amt = ifelse(condition & !is.na(condition), Amt, 0)) %>%
 mutate(condition = ifelse(is.na(condition),T,condition)) %>%
 filter(condition) %>%
 select(-condition)

首先创建一个与条件匹配的新列，然后根据该条件将其他列更改为0或NA。最后，取消选择新生成的列。（请注意，ifelse会将Date的班级更改为numeric，因此需要as.Date

  ID Location  startdate    enddate N  Loss_Date  Amt
1 A1     012A 2014-11-01 2014-12-31 2 2014-11-15 2200
2 A2     234B 2014-01-01 2014-08-31 0       <NA>    0
3 A3     012A 2015-10-01 2015-12-31 0       <NA>    0
4 A4     238C 2015-01-01 2015-12-31 2 2015-11-30  50

Answer 4

sqldf非常强大且易于阅读。检查此代码：

library(sqldf)
Output<-sqldf("
           SELECT L.*, r.N, r.Loss_Date, r.Amt
           FROM df1 as L
           LEFT JOIN df2 as r
           ON 
           L.ID=r.ID AND
              r.Loss_Date BETWEEN L.startdate AND L.enddate
           ORDER BY L.ID")

其中“L”代表df1（df1代表l），“r”代表df2（df2代表r）。

R：使用日期和时间以多个条件合并2个数据帧。 ID

4 个答案: