y<-data.frame(householdincome$State,householdincome$Income.Level,householdincome$Percent.of.Total)
factor(y$householdincome.Income.Level,levels = c("$15,000 to $19,999","$20,000 to $24,999","$200,000 or more"))
y[order(y$householdincome.Income.Level),]
我需要從 200,000 或更多開始訂購輸出,但是我的資料框輸出忽略該名稱的所有值,只列出我不尋找的列中的某些值。任何幫助將不勝感激。謝謝!
uj5u.com熱心網友回復:
這能解決您的問題嗎?
householdincome.State = c("AZ", "TX", "MT", "NY", "PA")
householdincome.Level = c("$15,000 to $19,999","$20,000 to $24,999","$200,000 or more", "$15,000 to $19,999","$20,000 to $24,999")
householdincome.Percent.of.Total = c(runif(5, 0, 100))
y <- data.frame(householdincome.State,
householdincome.Level,
householdincome.Percent.of.Total)
factor(y$householdincome.Level, levels = c("$15,000 to $19,999","$20,000 to $24,999","$200,000 or more"))
#> [1] $15,000 to $19,999 $20,000 to $24,999 $200,000 or more $15,000 to $19,999
#> [5] $20,000 to $24,999
#> Levels: $15,000 to $19,999 $20,000 to $24,999 $200,000 or more
y[order(y$householdincome.Level, decreasing = TRUE),]
#> householdincome.State householdincome.Level householdincome.Percent.of.Total
#> 3 MT $200,000 or more 95.45595
#> 2 TX $20,000 to $24,999 59.78140
#> 5 PA $20,000 to $24,999 71.03190
#> 1 AZ $15,000 to $19,999 48.62256
#> 4 NY $15,000 to $19,999 62.28109
由reprex 包(v2.0.1)于 2021 年 12 月 15 日創建
uj5u.com熱心網友回復:
或者你可以使用 [order(y$Level,decreasing=TRUE),]
householdincome<-data.frame("State" = c("London", "Reading", "Southampton"),
"Level" = c("$15,000 to $19,999","$20,000 to $24,999","$200,000 or more"),
"Percent.of.Total" = c(runif(3, 0, 100)))
y<-data.frame(householdincome)
y$Level<-factor(y$Level)
y[order(y$Level,decreasing=TRUE),]
State Level Percent.of.Total
3 Southampton $200,000 or more 10.37236
2 Reading $20,000 to $24,999 66.84539
1 London $15,000 to $19,999 98.65112
轉載請註明出處,本文鏈接:https://www.uj5u.com/ruanti/381269.html
上一篇:在R中建模的資料整理
下一篇:列中的逗號分隔值作為熊貓中的行
