Ale*_*xis 10 lookup r dataframe
首先,让我们从DataFrame 1(DF1)开始:
DF1 <- data.frame(c("06/19/2016", "06/20/2016", "06/21/2016", "06/22/2016",
"06/23/2016", "06/19/2016", "06/20/2016", "06/21/2016",
"06/22/2016", "06/23/2016"),
c(1, 1, 1, 1, 1, 2, 2, 2, 2, 2),
c(149, 150, 151, 152, 155, 84, 83, 80, 81, 97),
c(101, 102, 104, 107, 99, 55, 55, 56, 57, 58),
c("MTL", "MTL", "MTL", "MTL", "MTL", "NY", "NY",
"NY", "NY", "NY"))
colnames(DF1) <- c("date", "id", "sales", "cost", "city")
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我也有DataFrame 2(DF2):
DF2 <- data.frame(c("06/19/2016", "06/27/2016", "06/22/2016", "06/23/2016"),
c(1, 1, 2, 2),
c(9999, 8888, 777, 555),
c("LON", "LON", "QC", "QC"))
colnames(DF2) <- c("date", "id", "sales", "city")
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对于DF1中的每一行,我必须查看DF2中是否有一行具有相同的日期和ID.如果是,我必须用DF2中的值替换DF1中的值.
DF2的列总是比DF1少.如果列不在DF2中,我必须保留该特定列的DF1中的原始值.
最终输出是这样的:
results <- data.frame(c("06/19/2016", "06/20/2016", "06/21/2016", "06/22/2016",
"06/23/2016", "06/19/2016", "06/20/2016", "06/21/2016",
"06/22/2016", "06/23/2016"),
c(1, 1, 1, 1, 1, 2, 2, 2, 2, 2),
c(9999, 150, 151, 152, 155, 84, 83, 80, 777, 555),
c(101, 102, 104, 107, 99, 55, 55, 56, 57, 58),
c("LON", "MTL", "MTL", "MTL", "MTL", "NY", "NY",
"NY", "QC", "QC"))
colnames(results) <- c("date", "id", "sales", "cost", "city")
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你有什么建议吗?
Jaa*_*aap 21
您可以使用data.table -package的join功能:
library(data.table)
setDT(DF1)
setDT(DF2)
DF1[DF2, on = .(date, id), `:=` (city = i.city, sales = i.sales)]
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这使:
Run Code Online (Sandbox Code Playgroud)> DF1 date id sales cost city 1: 06/19/2016 1 9999 101 LON 2: 06/20/2016 1 150 102 MTL 3: 06/21/2016 1 151 104 MTL 4: 06/22/2016 1 152 107 MTL 5: 06/23/2016 1 155 99 MTL 6: 06/19/2016 2 84 55 NY 7: 06/20/2016 2 83 55 NY 8: 06/21/2016 2 80 56 NY 9: 06/22/2016 2 777 57 QC 10: 06/23/2016 2 555 58 QC
如果两个数据集中有许多列,则更容易使用mget
而不是键入所有列名称.对于问题中使用的数据,它看起来像:
DF1[DF2, on = .(date, id), names(DF2)[3:4] := mget(paste0("i.", names(DF2)[3:4]))]
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