Question

这与Are there more elegant ways to transform ragged data into a tidy dataframe

有关

为什么以下代码无效：

events = structure(list(date = structure(c(-714974, -714579, -717835), class = "Date"), 
    days = c(1, 6, 0.5), name = c("Intro to stats", "Stats Winter school", 
    "TidyR tools"), topics = c("probability|R", "R|regression|ggplot", 
    "tidyR|dplyr")), .Names = c("date", "days", "name", "topics"
), row.names = c(NA, -3L), class = "data.frame")

> newdf <- data.frame(topic=character(), days=character())
> for(i in 1:length(events$topics)){
+ xx = unlist(strsplit(events$topics[i],'\\|'))
+ for(j in 1:length(xx)){
+ yy = c(xx[j], events$days[i]/length(xx))
+ print(yy)
+ newdf=rbind(newdf, yy)
+ }
+ }
[1] "probability" "0.5"        
[1] "R"   "0.5"
[1] "R" "2"
[1] "regression" "2"         
[1] "ggplot" "2"     
[1] "tidyR" "0.25" 
[1] "dplyr" "0.25" 
There were 11 warnings (use warnings() to see them)
> newdf
  X.probability. X.0.5.
1    probability    0.5
2           <NA>    0.5
3           <NA>   <NA>
4           <NA>   <NA>
5           <NA>   <NA>
6           <NA>   <NA>
7           <NA>   <NA>
> 
> warnings()
Warning messages:
1: In `[<-.factor`(`*tmp*`, ri, value = structure(c(1L, NA ... :
  invalid factor level, NAs generated
2: In `[<-.factor`(`*tmp*`, ri, value = structure(c(1L, NA,  ... :
  invalid factor level, NAs generated
3: In `[<-.factor`(`*tmp*`, ri, value = structure(c(1L, 1L,  ... :
  invalid factor level, NAs generated
4: In `[<-.factor`(`*tmp*`, ri, value = structure(c(1L, NA,  ... :
  invalid factor level, NAs generated
5: In `[<-.factor`(`*tmp*`, ri, value = structure(c(1L, 1L,  ... :
  invalid factor level, NAs generated
6: In `[<-.factor`(`*tmp*`, ri, value = structure(c(1L, NA,  ... :
  invalid factor level, NAs generated
7: In `[<-.factor`(`*tmp*`, ri, value = structure(c(1L, 1L,  ... :
  invalid factor level, NAs generated
8: In `[<-.factor`(`*tmp*`, ri, value = structure(c(1L, NA,  ... :
  invalid factor level, NAs generated
9: In `[<-.factor`(`*tmp*`, ri, value = structure(c(1L, 1L,  ... :
  invalid factor level, NAs generated
10: In `[<-.factor`(`*tmp*`, ri, value = structure(c(1L, NA,  ... :
  invalid factor level, NAs generated
11: In `[<-.factor`(`*tmp*`, ri, value = structure(c(1L, 1L,  ... :
  invalid factor level, NAs generated
>

你可以，但是rbind没有用。错误在哪里以及如何纠正错误？谢谢你的帮助。

Answer 1

您可以尝试：

newdf <- data.frame(topic=character(), daysPerTopic=character(), stringsAsFactors=F)
for(i in 1:length(events$topics)){
xx = unlist(strsplit(events$topics[i],'\\|'))
for(j in 1:length(xx)){
yy = data.frame(topic=xx[j], daysPerTopic=events$days[i]/length(xx), stringsAsFactors=F)
newdf <- rbind(newdf, yy) 
 }
 }

 newdf
#        topic daysPerTopic
# 1 probability         0.50
# 2           R         0.50
# 3           R         2.00
# 4  regression         2.00
# 5      ggplot         2.00
# 6       tidyR         0.25
# 7       dplyr         0.25

或者

 op <- options(stringsAsFactors=F)  #set to F

 #Your code
 newdf <- data.frame(topic=character(), days=character())
 for(i in 1:length(events$topics)){
 xx = unlist(strsplit(events$topics[i],'\\|'))
 for(j in 1:length(xx)){
yy = c(xx[j], events$days[i]/length(xx))
print(yy)
newdf=rbind(newdf, yy)
 }
 }

 newdf
#  X.probability. X.0.5.
# 1    probability    0.5
# 2              R    0.5
# 3              R      2
# 4     regression      2
# 5         ggplot      2
# 6          tidyR   0.25
# 7          dplyr   0.25

 options(op) #et back to default

Answer 2

您是否尝试调试for循环？例如，通过添加print(class(yy)) print(str(newdf))，您会看到在第一次迭代后，两个newdf向量都成为因子。

# [1] "probability" "0.5"        
# [1] "character"
# 'data.frame':  0 obs. of  2 variables:
#   $ topic: Factor w/ 0 levels: 
#   $ days : Factor w/ 0 levels: 
#   NULL
# [1] "R"   "0.5"
# [1] "character"
# 'data.frame': 1 obs. of  2 variables:
#   $ X.probability.: Factor w/ 1 level "probability": 1
# $ X.0.5.        : Factor w/ 1 level "0.5": 1
# NULL
# [1] "R" "2"
# [1] "character"
# 'data.frame': 2 obs. of  2 variables:
#   $ X.probability.: Factor w/ 1 level "probability": 1 NA
# $ X.0.5.        : Factor w/ 1 level "0.5": 1 1

...

你会说＆＃34;但我将它们定义为character＆＃34;。没错，但是如果您阅读rbind文档，您会看到

对于cbind（rbind），忽略零长度（包括NULL）的向量除非结果为零行（列），否则为S兼容性。（零范围矩阵在S3中不会发生，在R中不会被忽略。）

rbind的另一个属性是它从data.frame继承了它的属性，而其中一个是stringsAsFactors == TRUE

这里发生的事情可以通过虚拟示例轻松说明，请考虑

temp <- data.frame(A = letters[1:3])
str(temp)
## 'data.frame':    3 obs. of  1 variable:
## $ A: Factor w/ 3 levels "a","b","c": 1 2 3

temp$A[3] <- "d"
## Warning message:
## In `[<-.factor`(`*tmp*`, 3, value = c(1L, 2L, NA)) :
##   invalid factor level, NA generated

temp$A
## [1] a    b    <NA>
## Levels: a b c

你可以在这里看到两件事：

data.frame自动将character类转换为因子
尝试将新关卡解析为factor向量时，会将其转换为NA并抛出您收到的确切错误

如@akrun所述，设置为options(stringsAsFactors=F)将解决您的问题

Answer 3

设置选项（stringsAsFactors = FALSE）并且您的代码应该按预期工作。结果中警告和NA的原因是因为隐式转换为因子以及newdf列和yy之间的类型不匹配，请参阅https://stackoverflow.com/a/1640729/1541036。

为了获得相同结果的更简洁方法，这里有一组使用data.table的解决方案

library(data.table)
events <- as.data.table(events)
events2 <- events[, list(topic=unlist(strsplit(topics, '|', fixed=TRUE))), by=c("date", "days", "name")]
events2[, probability := days / .N, by=name]

为什么rbind会发出警告

3 个答案: