1

我正在尝试使用 tidyr 扩展函数，但我想传入我自己的特征名称向量以用于关键参数。

例如，默认用法是

test<-data.frame(id=c(1,1,2,2),
             feat=c("feat1", "feat2", "feat1", "feat2"),
             value = c(10,20, 1000, 2000))
test %>% spread(key = feat, value = value, fill = 0)
  id feat1 feat2
1  1    10    20
2  2  1000  2000

我想传入我自己的特征字符串向量作为键，就像这样。

featlist<-c("feat1", "feat2", "feat3")
test %>% spread(key = featlist, value = value, fill = 0)
#desired output
  id feat1 feat2 feat3
1  1    10    20     0
2  2  1000  2000     0
#Error output
Error: `var` must evaluate to a single number or a column name, not a character vector
#Trying spread_
test %>% spread_(key = featlist, value = "value", fill = 0)
Error: Only strings can be converted to symbols
4

2 回答 2

4

只需将专长列设置为级别设置为的因素，featlist然后将drop参数设置FALSE为：

test<-data.frame(id=c(1,1,2,2),
                 feat=c("feat1", "feat2", "feat1", "feat2"),
                 value = c(10,20, 1000, 2000))

featlist<-c("feat1", "feat2", "feat3")
test$feat <- factor(test$feat, levels = featlist)

test %>% spread(key = feat, value = value, fill = 0, drop = FALSE)

结果是：

  id feat1 feat2 feat3
1  1    10    20     0
2  2  1000  2000     0
于 2018-04-19T22:39:32.637 回答
1

不幸的是，tidyr::spread不允许使用你自己的vectoraskey但幸运的是expand.grid为你提供了一个选项来使用你自己的vector并data.frame在调用spread函数之前进行扩展。

library(tidyverse)
expand.grid(id=unique(test$id), feat = featlist) %>% #creates all combinations
  mutate(feat = as.character(feat)) %>%  
  left_join(test, by=c("id", "feat")) %>%      #Join with actual dataframe
  spread(key=feat, value = value, fill = 0)

#  id feat1 feat2 feat3
#1  1    10    20     0
#2  2  1000  2000     0

数据：

test<-data.frame(id=c(1,1,2,2),
                 feat=c("feat1", "feat2", "feat1", "feat2"),
                 value = c(10,20, 1000, 2000), stringsAsFactors = FALSE)

featlist<-c("feat1", "feat2", "feat3")
于 2018-04-19T22:49:20.097 回答