Question

我有一行代码需要每次运行12次并更换字符串。

unique1 <- unique(master[c("Country_University_1", "City_University_1", "Latitude_city_1", "Longitude_city_1")])

通过将每个列名末尾的数字递增1，这将生成12个单独的数据帧，然后我将与rbind

绑定在一起

我的目标是编写一个函数，它将列名和的最后一个字符作为对象名称，并像x <- x + 1一样递增1，并在12处停止。我一直在绝望地尝试写一个看起来像

的函数

x <- c("Country_University_1", "City_University_1", "Latitude_city_1", "Longitude_city_1")
for (i in 1:12){
  gsub("i", nchar(x[1:4]-1 ) x[1:4])
  print(x)
}

但我仍然不熟悉R语法。任何建议将不胜感激

完全写出，该函数将执行以下操作：

unique1 <- unique(master[c("Country_University_1", "City_University_1", "Latitude_city_1", "Longitude_city_1")])
unique2 <- unique(master[c("Country_University_2", "City_University_2", "Latitude_city_2", "Longitude_city_2")])
unique3 <- unique(master[c("Country_University_3", "City_University_3", "Latitude_city_3", "Longitude_city_3")])
unique4 <- unique(master[c("Country_University_4", "City_University_4", "Latitude_city_4", "Longitude_city_4")])
unique5 <- unique(master[c("Country_University_5", "City_University_5", "Latitude_city_5", "Longitude_city_5")])
unique6 <- unique(master[c("Country_University_6", "City_University_6", "Latitude_city_6", "Longitude_city_6")])
unique7 <- unique(master[c("Country_University_7", "City_University_7", "Latitude_city_7", "Longitude_city_7")])
unique8 <- unique(master[c("Country_University_8", "City_University_8", "Latitude_city_8", "Longitude_city_8")])
unique9 <- unique(master[c("Country_University_9", "City_University_9", "Latitude_city_9", "Longitude_city_9")])
unique10 <- unique(master[c("Country_University_10", "City_University_10", "Latitude_city_10", "Longitude_city_10")])
unique11 <- unique(master[c("Country_University_11", "City_University_11", "Latitude_city_11", "Longitude_city_11")])
unique12 <- unique(master[c("Country_University_12", "City_University_12", "Latitude_city_12", "Longitude_city_12")])

output <- rbind(unique1, unique2, unique3, unique4, unique5, unique6, unique7, unique8, unique9, unique10, unique11, unique12)

Answer 1

重做以匹配请求。

Base = c("Country_University_", "City_University_", "Latitude_city_", "Longitude_city_")
output = rep(NULL,4)
for(i in 1:12) {
    ColNames = sapply(Base, function(s) { paste(s, i, sep="") })
    output=rbind(output, unique(master[ColNames]))
}

此代码未经过测试，因为我们没有掌握，但输出应该是您想要的。

Answer 2

您需要在sub使用一些正则表达式。看起来你想用越来越多的数字替换字符串末尾的数字，你可以检查一下，模式$中匹配字符串结尾的位置，因此\\d+$匹配数字末尾字符串：

for (i in 1:2) { print(sub("\\d+$", i, x)) }

#[1] "Country_University_1" "City_University_1"    "Latitude_city_1"     
#[4] "Longitude_city_1"    
#[1] "Country_University_2" "City_University_2"    "Latitude_city_2"     
#[4] "Longitude_city_2"

根据您的更新，您可能需要这样的内容：

# get the number at the end of column names
number = sub(".*_(\\d+)$", "\\1", names(master))

# split the data frame by the number, rename each sub data frame and bind them together
do.call(rbind, lapply(split.default(master, number), 
                      function(sdf) setNames(sdf, sub("_\\d+$", "", names(sdf)))))

Answer 3

我们可以使用melt中的data.table轻松完成此操作，这可能需要多个measure patterns

library(data.table)
melt(setDT(master), measure = patterns("^Country", "^City", "^Latitude", "^Longitude"), 
  value.name = c("Country_University", "City_University", "Latitude_city", "Longitude_city")
         ][, variable := NULL][]

字符串替换中增加1

3 个答案: