【发布时间】:2018-01-21 05:13:14
【问题描述】:
作为我在下面附加的图像和代码,我有一组交易数据,每一行都有它的行业名称。
可重现的示例数据:
structure(list(Date = c(1201, 1201, 1201, 1201, 1201, 1201, 1201,
1201, 1201, 1201), Sex = c("Male", "Male", "Female", "Male",
"Male", "Female", "Male", "Female", "Male", "Male"), Age = c(10,
15, 20, 15, 40, 50, 20, 30, 50, 20), City = c("Pheonix", "Atlanta",
"Las Vegas", "Las Vegas", "Denver", "Pheonix", "Atlanta", "Las Vegas",
"Las Vegas", "Minneapolis"), State = c("Arizona", "Georgia",
"Nevada", "Nevada", "Colorado", "Arizona", "Georgia", "Nevada",
"Nevada", "Minesota"), Industry = c("food", "furniture", "clothes",
"transportation", "leisure", "food", "furniture", "food", "transportation",
"furniture"), `no.of users` = c(48, 50, 83, 111, 186, 196.7,
230.4, 264.1, 297.8, 331.5), `no. of approval cases` = c(48,
21, 25, 48, 70, 63.7, 70.8, 77.9, 85, 92.1), `Total spending` = c(1541000,
512000, 1757000, 1117000, 1740500, 1634700, 1735100, 1835500,
1935900, 2036300)), .Names = c("Date", "Sex", "Age", "City",
"State", "Industry", "no.of users", "no. of approval cases",
"Total spending"), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA,
-10L))
我想做的是仅对行业名称与条件匹配的行进行子集化,并将其导出为单独的 csv 文件,文件名与其行业名称相对应。 例如,假设我有一个列表
current_industry<-c("food","furniture")
首先,我想对行业名称匹配"food"的行进行子集化,然后将其保存到csv files name 'food',然后将行业名称匹配到"furniture"的以下行子集化并保存为单独的@ 987654327@名称'家具'。
我编写了 for 循环来执行此操作,但这不起作用,因为以下子集数据会预先替换那个。
for(i in current_industry){
write.csv(subset(masterset, industry == i), "i.csv")}
【问题讨论】:
-
请分享
masterset作为可重复的示例。使用dput分享,以便我们测试如何解决您的问题。 -
你需要像
paste0(i, ".csv")这样的东西作为 write.csv 的第二个参数 -
抱歉,忘记分享示例。我编辑以分享可重复的示例。谢谢。
-
第二个参数是什么意思?你能更详细地解释一下吗?谢谢..!