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,而且我的想法已经用完了。
以下是Google云端硬盘上的dataset。
以下是我需要做的事情:
到目前为止,这是我所做的代码:
# 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), ]
运行此代码后,拼写错误的单词(“Ackland”)不会像我预期的那样改为奥克兰。 此图中显示的突出显示的行应该更改为奥克兰:
答案 0 :(得分:2)
解决您的问题崩溃因素水平&#39; Ackland&#39;和奥克兰&#39; (并且假设你想要力量$ City是/仍然是一个因素):
一种方法是将levels()函数传递给一个命名列表,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)作为一般例子。
然而,就像抬头一样,请注意数据集中Ackland和Auckland字符串背后的额外空间:
# 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$City)
[1] "Ackland " "Auckland " "Sydney" "Wellington"
在考虑空格后,将命名列表传递给levels()即可运行:
levels(power$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$City)
[1] "Auckland" "Sydney" "Wellington"
你现在有3个级别而不是4级,注意这也处理了级别标签中的空格
子集仅包含相关城市
subpower <- power[which(power$City == c("Auckland", "Wellington")), ]
你也可以通过子集来排除负值,极值等......
注意:我唯一真正的贡献就是抓住额外的空间,自己解决类似的问题Aaron's 回答非常有帮助。希望这有帮助!