首页 文章

将函数应用于匹配多行的组

提问于
浏览 次
1

我正在尝试向我的df添加一个新列，这只是我的函数hardfunct应用于'values'，其中行是'hardness' . 然后，我希望该值填充该列中匹配'site'和'dates'的所有行 . 如何填充其余行？我尝试过使用summary，rowwise和mutate . 样本数据如下 .

site=c(rep("River A",4),rep("River B",4))
dates=as.Date(c("01/01/2001","01/01/2001","01/01/2001","01/01/2001","05/08/2001","05/08/2001","05/08/2001","05/08/2001"),  format = "%m/%d/%Y")
param=c("lead","hardness","mercury","cadmium","lead","hardness","mercury","cadmium")
value=c("0.2","45","0.9","1.2","0.5","1800","0.6","0.8")

df=data.frame(site,param,dates,value)

hardfunct=function(x){
if (x>=400) {
print(400)
} else if (x<=25) {
print(25)
} else {
return(x)}
}

#######Trying to use group_by and mutate

df %>% group_by(site,dates) %>% 
mutate(New_Hardness=sapply(df[df$param=="hardness","value"],hardfunct))

这是新列的数据框应该是什么样子

site      param     dates     value New_Hardness
River A   lead      1/1/2001    0.2   45
River A   hardness  1/1/2001    45    45
River A   mercury   1/1/2001    0.9   45
River A   cadmium   1/1/2001    1.2   45
River B   lead      5/8/2001    0.5   400
River B   hardness  5/8/2001    1800  400
River B   mercury   5/8/2001    0.6   400
River B   cadmium   5/8/2001    0.8   400
r dplyr mutate

2 回答

  • 2

    在基数R中，您可以使用拆分/应用/组合策略 .

    请注意， pmax 和 pmin 的想法是@Frank's .

    sp <- split(df, list(df$site, df$dates))
sp <- sp[sapply(sp, function(x) nrow(x) != 0)]
newdf <- lapply(sp, function(DF) {
    DF$New_Hardness <- pmax(25, pmin(400, DF$value[DF$param == "hardness"]))
    DF
})

rm(sp)    # tidy up

newdf <- do.call(rbind, newdf)
row.names(newdf) <- NULL

newdf
#     site    param      dates  value New_Hardness
#1 River A     lead 2001-01-01    0.2           45
#2 River A hardness 2001-01-01   45.0           45
#3 River A  mercury 2001-01-01    0.9           45
#4 River A  cadmium 2001-01-01    1.2           45
#5 River B     lead 2001-05-08    0.5          400
#6 River B hardness 2001-05-08 1800.0          400
#7 River B  mercury 2001-05-08    0.6          400
#8 River B  cadmium 2001-05-08    0.8          400
    回复于
  • 1
    site=c(rep("River A",4),rep("River B",4))
dates=as.Date(c("01/01/2001","01/01/2001","01/01/2001","01/01/2001","05/08/2001","05/08/2001","05/08/2001","05/08/2001"),  format = "%m/%d/%Y")
param=c("lead","hardness","mercury","cadmium","lead","hardness","mercury","cadmium")
value=c("0.2","45","0.9","1.2","0.5","1800","0.6","0.8")

df=data.frame(site,param,dates,value, stringsAsFactors = F)

hardfunct=function(x){
  if (x>=400) {
    return(400)
  } else if (x<=25) {
    return(25)
  } else {
    return(x)}
}

library(dplyr)

df %>%
  group_by(site, dates) %>%
  mutate(New = hardfunct(as.numeric(value[param == "hardness"]))) %>%
  ungroup()

# # A tibble: 8 x 5
#   site    param    dates      value   New
#   <chr>   <chr>    <date>     <chr> <dbl>
# 1 River A lead     2001-01-01 0.2      45
# 2 River A hardness 2001-01-01 45       45
# 3 River A mercury  2001-01-01 0.9      45
# 4 River A cadmium  2001-01-01 1.2      45
# 5 River B lead     2001-05-08 0.5     400
# 6 River B hardness 2001-05-08 1800    400
# 7 River B mercury  2001-05-08 0.6     400
# 8 River B cadmium  2001-05-08 0.8     400

    请注意，您必须将函数中的 print 更改为 return ，否则您还需要在数据帧输出之前获取打印值 .

    另请注意，您需要具有字符变量而不是因子，因为应用于因子的 as.numeric 将为您提供与您期望的不同的数字 .

    回复于

相关问题