我有2个数据集
> df2<-data.frame(name=c("A","B","C"),F1=c(5,8,9),F2=c(3,8,9),F3=c(1,2,3))
name F1 F2 F3
1 A 5 3 1
2 B 8 8 2
3 C 9 9 3
> df1<-data.frame(name=c("C","C","A","B"),F1=c(1,5,8,9),F2=c(1,5,8,9),F3=c(1,5,8,9))
name F1 F2 F3
1 C 1 1 1
2 C 5 5 5
3 A 8 8 8
4 B 9 9 9
我想用通用名称（例如C / C，C / C，A / A，B / B）将df1除以df2。这只是我的数据集的一个示例，谢谢是250x50。
答案 0 :(得分：2)
请注意，如果您的任何真实数据名称包含undefined reference to `curlpp::Cleanup::Cleanup()'
path_service.cpp:(.text+0xf9): undefined reference to `curlpp::Easy::Easy()'
path_service.cpp:(.text+0x12a): undefined reference to `curlpp::Easy::setOpt(curlpp::OptionBase*)'
path_service.cpp:(.text+0x15b): undefined reference to `curlpp::Easy::setOpt(curlpp::OptionBase*)'
path_service.cpp:(.text+0x18c): undefined reference to `curlpp::Easy::setOpt(curlpp::OptionBase*)'
path_service.cpp:(.text+0x1d3): undefined reference to `curlpp::Easy::setOpt(curlpp::OptionBase*)'
path_service.cpp:(.text+0x218): undefined reference to `curlpp::Easy::setOpt(curlpp::OptionBase*)'
path_service.cpp:(.text+0x227): undefined reference to `curlpp::Easy::perform()'
path_service.cpp:(.text+0x298): undefined reference to `curlpp::Easy::~Easy()'
path_service.cpp:(.text+0x2a7): undefined reference to `curlpp::Cleanup::~Cleanup()'
path_service.cpp:(.text+0x384): undefined reference to `curlpp::Easy::~Easy()'
path_service.cpp:(.text+0x398): undefined reference to `curlpp::Cleanup::~Cleanup()'
或".x"，则此操作将无效。
".y"答案 1 :(得分：2)
我们可以使用data.table的联接和除法
library(data.table)
nm1 <- names(df1)[-1]
nm2 <- paste0('i.', nm1)
setDT(df1)[df2, (nm1) := Map(`/`, mget(nm1), mget(nm2)), on = .(name)]
df1
# name F1 F2 F3
#1: C 0.1111111 0.1111111 0.3333333
#2: C 0.5555556 0.5555556 1.6666667
#3: A 1.6000000 2.6666667 8.0000000
#4: B 1.1250000 1.1250000 4.5000000
答案 2 :(得分：2)
这是基本的R解决方案：
ind = match(df1$name, df2$name)
data.frame("name" = df1$name, df1[,-1] / df2[ind,-1])
# name F1 F2 F3
# 1 C 0.1111111 0.1111111 0.3333333
# 2 C 0.5555556 0.5555556 1.6666667
# 3 A 1.6000000 2.6666667 8.0000000
# 4 B 1.1250000 1.1250000 4.5000000