Question

Material    DocDate    Name  Address    Unit    Price
1258486   3/17/2017   FEHLIG BROS BOX    asd     8.95
1258486   5/11/2017   FEHLIG BROS BOX    asd     9.5
1258486   12/11/2017  FEHLIG BROS_BOX    asd     10.5
1250000   12/20/2017  Krones ALPHA       afg     11.5

我有一个以上数据框。我需要像下面这样基于日期（3/17/2017）出现的框架。所以我需要下面的输出

Material         Name/address/Unit Price
1258486     FEHLIG BROS BOX/asd/8.95/9.5/10.5
1250000     Krones/ALPHA/afg/11.5

Answer 1

使用data.table可以尝试

df <- read.table(stringsAsFactors = FALSE, header = TRUE,
                 text ="Material DocDate Name  Address Unit  Price
                 1258486   3/17/2017  FEHLIG BROS_BOX     asd     8.95
                 1258486   5/11/2017  FEHLIG BROS_BOX     asd     9.5
                 1258486   12/11/2017  FEHLIG BROS_BOX    asd     10.5
                 1250000   12/20/2017  Krones ALPHA       afg     11.5
                 ")
df$DocDate <- as.Date(df$DocDate,'%m/%d/%Y')
library(data.table)
setDT(df)[,.(newVar = paste(Name, Address, Unit, paste(.SD$Price,collapse = "/"), sep = "/") )
          ,by = Material][,.(newVar = newVar[1]), Material]

#returns
   Material                            newVar
1:  1258486 FEHLIG/BROS_BOX/asd/8.95/9.5/10.5
2:  1250000             Krones/ALPHA/afg/11.5

Answer 2

这是使用dplyr的替代方法。首先是示例数据：

data <- data.frame(stringsAsFactors=FALSE,
                   Material   = c(1258486L, 1258486L),
                   DocDate    = c("3/17/2017", "5/11/2017"),
                   Name       = c("FEHLIG BROS BOX", "FEHLIG BROS BOX"),
                   Address    = c("asd", "asd"),
                   Unit_Price = c(8.95, 9.5))

然后是获取您答案的一组步骤。（顺便说一句，我相信，如果有多个Material行共享相同的“最早日期”，那么到目前为止提供的所有解决方案都会为您提供多行输出。 Unit_Price == min(Unit_Price)（如果在这里有合理的平局）。

filter

（编辑：代码中的固定错字）

Answer 3

根据您对问题的更改进行的完整编辑：

# create example data (notice this differs slightly from your table above)
df <- read.csv(stringsAsFactors = FALSE, header = TRUE,
                 text ="Material, DocDate, Name, Address, UnitPrice
                        1258486, 3/17/2017, FEHLIG BROS BOX, asd, 8.95
                        1258486, 5/11/2017, FEHLIG BROS BOX, asd, 9.50
                        1258486, 12/11/2017, FEHLIG BROS_BOX, asd, 10.5
                        1250000, 12/20/2017, Krones ALPHA, afg, 11.5")

# let's use data.table
library(data.table)
df_orig <- as.data.table(df)
df_orig[ , DocDate := as.Date(DocDate,format="%m/%d/%Y")][order(DocDate)]

# create one string per Name-Material pair
df_intermed <- df_orig[ , .(newvar = paste(Name[1], Address[1], paste(UnitPrice, collapse="/"), sep="/")), by=.(Material, Name)]

# aggregate those strings across Names, so one row per Material
df_final <- df_intermed[ , .(newvar = paste(newvar, collapse=",")), by=Material]

将一个数据帧中的列合并为新数据帧，然后进行过滤

3 个答案: