使用子集识别和纠正数据集中的拼写错误
Identifying and correcting typos in a data set using subset
我有一个数据集：（从下面可以看出link）
https://drive.google.com/file/d/0B4Mldbnr1-avMDIxYmZLSnRfUDA/view?usp=sharing 并且我想使用子集和级别函数进行数据校正。这是我一直在尝试应用但似乎不起作用的方法：
# Setting working directory
setwd("F:/Intro Data Science/Assignment Part B/Assignment Part B-20170902")
plot.new()
options(digits=2)
# 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"
我真的需要你们的帮助。这本来应该很容易，因为我完全按照讲座中给出的内容进行操作，但是当我 运行 代码时它什么也没做。
感谢你的帮助
纳尔逊
请求的输出：
> dput(head(power, 30))
structure(list(Area = c(144.38, 176.83, 268.71, 208.67, 123.61,
199.3, 109.46, 183.28, 110.61, 146.91, 77.451, 232.65, 270.94,
49.191, 234.5, 280.93, 192.18, 95.918, 230.74, 72.698, 129.26,
110.76, 199.44, 129.75, 146.8, 287.97, 162.1, 249.03, 159.3,
272.51), City = c("Auckland ", "Auckland ", "Auckland ", "Auckland ",
"Auckland ", "Auckland ", "Auckland ", "Auckland ", "Auckland ",
"Auckland ", "Auckland ", "Auckland ", "Auckland ", "Auckland ",
"Auckland ", "Auckland ", "Auckland ", "Ackland ", "Auckland ",
"Auckland ", "Auckland ", "Auckland ", "Auckland ", "Auckland ",
"Auckland ", "Auckland ", "Auckland ", "Auckland ", "Auckland ",
"Auckland "), P.Winter = c(1684.9, 1926.7, 2026.9, 1938.1, 1579.9,
1991.4, 1572.5, 1691.2, 1684.2, 1743.6, 1234.6, 2043, 1986.7,
1259.7, 1870.4, 2115.6, 18000, 1452, 1936.2, 1430.2, 1587.3,
1614.3, 1993.2, 1746.4, 1807.6, 2009.4, 1859.1, 1985.5, 1909.4,
1892.7), P.Summer = c(1194.5, 1487.3, 1737.3, -158, 1148.1, 1445.8,
885.77, 1393, 1191.5, 1149.9, 813.38, 1623.8, 1708, 874.48, 1635.7,
1826.1, 1596.6, 793.71, 1668.8, 905.6, 1227.3, 938.38, 1523.1,
1012.6, 1122.8, 1829.5, 1223.3, 1653.2, 1175.5, 1882)), .Names = c("Area",
"City", "P.Winter", "P.Summer"), row.names = c(NA, 30L), class = "data.frame")
我相信你想要的功能是droplevels。
先补个资料。
set.seed(5295) # make the results reproducible
cities <- factor(sample(c("Ackland", "Auckland", "Wellington", "Sidney"), 100, TRUE))
power <- data.frame(City = cities)
现在是代码，从你的开始。
power$City[power$City == "Ackland"] <- "Auckland"
power$City <- droplevels(power$City)
levels(power$City) # check if it worked
#[1] "Auckland" "Sidney" "Wellington"
编辑。
在看到 dput(head(power, 30)) 的输出后，解决方案就变得显而易见了。 City 列属于 class character，而不是 factor，并且没有值 "Ackland" 或 "Auckland"，它们有一个尾随的白色 space 搞砸了。所以我们需要做的就是删除 "Ackland " 并删除尾随的白色 spaces.
str(power)
#'data.frame': 30 obs. of 4 variables:
# $ Area : num 144 177 269 209 124 ...
# $ City : chr "Auckland " "Auckland " "Auckland " "Auckland " ...
# $ P.Winter: num 1685 1927 2027 1938 1580 ...
# $ P.Summer: num 1194 1487 1737 -158 1148 ...
which(power$City == "Ackland ") # note the white space
#[1] 18
which(power$City == "Auckland ") # note the white space
# [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 19 20 21 22 23 24 25 26
#[26] 27 28 29 30
# remove the value "Ackland ", with white space
power$City[power$City == "Ackland "] <- "Auckland"
power$City <- trimws(power$City) # remove white spaces from all of them
并且没有任何列消失，只是 运行 str(power) 可以看到它。
我有一个数据集：（从下面可以看出link） https://drive.google.com/file/d/0B4Mldbnr1-avMDIxYmZLSnRfUDA/view?usp=sharing 并且我想使用子集和级别函数进行数据校正。这是我一直在尝试应用但似乎不起作用的方法：
# Setting working directory
setwd("F:/Intro Data Science/Assignment Part B/Assignment Part B-20170902")
plot.new()
options(digits=2)
# 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"
我真的需要你们的帮助。这本来应该很容易，因为我完全按照讲座中给出的内容进行操作，但是当我 运行 代码时它什么也没做。 感谢你的帮助 纳尔逊
请求的输出：
> dput(head(power, 30))
structure(list(Area = c(144.38, 176.83, 268.71, 208.67, 123.61,
199.3, 109.46, 183.28, 110.61, 146.91, 77.451, 232.65, 270.94,
49.191, 234.5, 280.93, 192.18, 95.918, 230.74, 72.698, 129.26,
110.76, 199.44, 129.75, 146.8, 287.97, 162.1, 249.03, 159.3,
272.51), City = c("Auckland ", "Auckland ", "Auckland ", "Auckland ",
"Auckland ", "Auckland ", "Auckland ", "Auckland ", "Auckland ",
"Auckland ", "Auckland ", "Auckland ", "Auckland ", "Auckland ",
"Auckland ", "Auckland ", "Auckland ", "Ackland ", "Auckland ",
"Auckland ", "Auckland ", "Auckland ", "Auckland ", "Auckland ",
"Auckland ", "Auckland ", "Auckland ", "Auckland ", "Auckland ",
"Auckland "), P.Winter = c(1684.9, 1926.7, 2026.9, 1938.1, 1579.9,
1991.4, 1572.5, 1691.2, 1684.2, 1743.6, 1234.6, 2043, 1986.7,
1259.7, 1870.4, 2115.6, 18000, 1452, 1936.2, 1430.2, 1587.3,
1614.3, 1993.2, 1746.4, 1807.6, 2009.4, 1859.1, 1985.5, 1909.4,
1892.7), P.Summer = c(1194.5, 1487.3, 1737.3, -158, 1148.1, 1445.8,
885.77, 1393, 1191.5, 1149.9, 813.38, 1623.8, 1708, 874.48, 1635.7,
1826.1, 1596.6, 793.71, 1668.8, 905.6, 1227.3, 938.38, 1523.1,
1012.6, 1122.8, 1829.5, 1223.3, 1653.2, 1175.5, 1882)), .Names = c("Area",
"City", "P.Winter", "P.Summer"), row.names = c(NA, 30L), class = "data.frame")
我相信你想要的功能是droplevels。
先补个资料。
set.seed(5295) # make the results reproducible
cities <- factor(sample(c("Ackland", "Auckland", "Wellington", "Sidney"), 100, TRUE))
power <- data.frame(City = cities)
现在是代码，从你的开始。
power$City[power$City == "Ackland"] <- "Auckland"
power$City <- droplevels(power$City)
levels(power$City) # check if it worked
#[1] "Auckland" "Sidney" "Wellington"
编辑。
在看到 dput(head(power, 30)) 的输出后，解决方案就变得显而易见了。 City 列属于 class character，而不是 factor，并且没有值 "Ackland" 或 "Auckland"，它们有一个尾随的白色 space 搞砸了。所以我们需要做的就是删除 "Ackland " 并删除尾随的白色 spaces.
str(power)
#'data.frame': 30 obs. of 4 variables:
# $ Area : num 144 177 269 209 124 ...
# $ City : chr "Auckland " "Auckland " "Auckland " "Auckland " ...
# $ P.Winter: num 1685 1927 2027 1938 1580 ...
# $ P.Summer: num 1194 1487 1737 -158 1148 ...
which(power$City == "Ackland ") # note the white space
#[1] 18
which(power$City == "Auckland ") # note the white space
# [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 19 20 21 22 23 24 25 26
#[26] 27 28 29 30
# remove the value "Ackland ", with white space
power$City[power$City == "Ackland "] <- "Auckland"
power$City <- trimws(power$City) # remove white spaces from all of them
并且没有任何列消失，只是 运行 str(power) 可以看到它。