变异和 ifelse() 失败，因为列中存在 NA

Question

我在尝试使用 ifelse 创建新列时遇到问题。非常相似的问题是这个dplyr error: strange issue when combining group_by, mutate and ifelse. Is it a bug?

set.seed(101)
time =sort(runif(10,0,10))  
group=rep(c(1,2),each=5)
az=c(sort(runif(5,-1,1),decreasing = T),sort(runif(5,-1,0.2),decreasing = T))

df <- data.frame(time,az,group)

#       time          az group
#1  0.4382482  0.86326886     1
#2  2.4985572  0.75959146     1
#3  3.0005483  0.46394519     1
#4  3.3346714  0.41374948     1
#5  3.7219838 -0.08975881     1
#6  5.4582855 -0.01547669     2
#7  5.8486663 -0.29161632     2
#8  6.2201196 -0.50599980     2
#9  6.5769040 -0.73105782     2
#10 7.0968402 -0.95366733     2

在df 中，我正在尝试条件变异clas 列。但是，由于sw_time 内部有NA，因此所有clas 列也变为NA，其中group 1 通常应该是nrm。

df1 <- df%>%
  group_by(group)%>%
  mutate(sw_time=abs(time[which(az<=0.8)[1]]-time[which(az>0)[1]]))%>%
  mutate(clas=as.numeric(ifelse(sw_time<3,"nrm","abn")))

Source: local data frame [10 x 5]
Groups: group [2]

        time          az group  sw_time  clas
       (dbl)       (dbl) (dbl)    (dbl) (dbl)
1  0.4382482  0.86326886     1 2.060309    NA
2  2.4985572  0.75959146     1 2.060309    NA
3  3.0005483  0.46394519     1 2.060309    NA
4  3.3346714  0.41374948     1 2.060309    NA
5  3.7219838 -0.08975881     1 2.060309    NA
6  5.4582855 -0.01547669     2       NA    NA
7  5.8486663 -0.29161632     2       NA    NA
8  6.2201196 -0.50599980     2       NA    NA
9  6.5769040 -0.73105782     2       NA    NA
10 7.0968402 -0.95366733     2       NA    NA

提前感谢您的行动！

Answer 1

通过将character 类转换为numeric，将得到NA。相反，我们可能需要一个 factor 类来强制转换为 numeric

df %>%
    group_by(group)%>%
     mutate(sw_time=abs(time[which(az<=0.8)[1]]-time[which(az>0)[1]]),
            clas=as.integer(factor(ifelse(sw_time<3,"nrm","abn"))))

如果我们只对获取 'nrm'、'abn' 感兴趣，只需删除 as.integer(factor 包装

df%>%
  group_by(group)%>%
  mutate(sw_time=abs(time[which(az<=0.8)[1]]-time[which(az>0)[1]]),
          clas=ifelse(sw_time<3,"nrm","abn"))
#        time          az group  sw_time  clas
#       <dbl>       <dbl> <dbl>    <dbl> <chr>
#1  0.4382482  0.86326886     1 2.060309   nrm
#2  2.4985572  0.75959146     1 2.060309   nrm
#3  3.0005483  0.46394519     1 2.060309   nrm
#4  3.3346714  0.41374948     1 2.060309   nrm
#5  3.7219838 -0.08975881     1 2.060309   nrm
#6  5.4582855 -0.01547669     2       NA  <NA>
#7  5.8486663 -0.29161632     2       NA  <NA>
#8  6.2201196 -0.50599980     2       NA  <NA>
#9  6.5769040 -0.73105782     2       NA  <NA>
#10 7.0968402 -0.95366733     2       NA  <NA>

---

我们也可以使用data.table

library(data.table)
setDT(df)[, c("sw_time", "clas") := {
           v1 <- abs(time[which(az <= 0.8)[1]] - time[which(az > 0)[1]])
          .(v1 , c("abn", "nrm")[(v1 < 3) + 1]) },
                      by = group]

---

如果最终输出不涉及'nrm'、'abn'，我们就不需要ifelse部分。我们可以直接使用as.integer(sw_time <3)