table 中的数据清理和拼错单词

Data cleaning and misspelled words in a table

我有这个 CSV 数据集,我需要创建一个函数来执行数据清理,但仍然无法正常工作,我 运行 想不通了。

这是 Google 驱动器上的 dataset

这是我需要做的:

到目前为止,这是我完成的代码:

# Reading data set
installed.packages("lubridate")
library(lubridate)

# Reading data set
power <- read.csv("data set 6.csv", na.strings="")

# SUBSETTING
Area <- as.numeric(power$Area)
City <- as.character(power$City)
P.Winter <- as.numeric(power$P.Winter)
P.Summer <- as.numeric(power$P.Summer)

#Data Cleaning
levels(power$City) <- c(levels(power$City), "Auckland")
power$City[power$City == "Ackland"] <- "Auckland"

#Removing irrelevant data (only houses in Auckland and Wellington are considered)
power$City <- power$City[-c(496,499), ]

在我 运行 这段代码之后,拼错的单词 ("Ackland") 并没有像我预期的那样变成奥克兰。 如图所示,突出显示的行应该更改为奥克兰:

为了解决你的问题,折叠因子水平 'Ackland' 和 'Auckland'(并且假设你想要 power$City 到 be/remain 一个因子):

一种方法是向 levels() 函数传递一个命名列表,每个名称都是所需级别的正确标签(在您的情况下是数据集中城市的正确名称)请参阅: Cleaning up factor levels (collapsing multiple levels/labels) 举个一般的例子。

但是,请注意,注意数据集中 Ackland 和 Auckland 字符串后面的额外 space:

    # first view classes to confirm power$City is a factor
     > apply(power, class)     # --> or is.factor(power$City) will work to
        Area      City  P.Winter  P.Summer 
    "numeric"  "factor" "numeric" "numeric" 

    # Notice spaces behind "Ackland " and "Auckland "
     > levels(power$City)
    [1] "Ackland "   "Auckland "  "Sydney"     "Wellington"

将命名列表传递给 levels() 可以在您考虑 space 后起作用:

    levels(power$City) <-  list(Auckland = c("Ackland ", "Auckland "), Sydney = c("Sydney"), Wellington = c("Wellington"))

    # Now only three factor levels (notice this also took care of the extra spaces)
      > levels(power$City)
     [1] "Auckland"   "Sydney"     "Wellington"

你现在有 3 个级别而不是 4 个,请注意这也处理了级别标签 space 中的

子集仅包含相关城市

       subpower <- power[which(power$City == c("Auckland", "Wellington")), ]

您还可以进行子集化以排除负值、极值等...

注意:我在这里唯一真正的贡献是抓住了额外的 spaces,为了自己解决类似的问题,我发现 Aaron's answer 非常有帮助。希望这对您有所帮助!