R中的混合排序
我在重新调整字母数字列时遇到了一些问题，正在寻找一些建议 考虑以下几点：R中的混合排序,r,sorting,mixed,R,Sorting,Mixed,我在重新调整字母数字列时遇到了一些问题，正在寻找一些建议 考虑以下几点： structure(list(Company = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 3L, 3L), .Label = c("ABC Inc", "ACME", "Handy Co"), class = "factor"), Quarter = structure(c(2L, 3L, 4L, 5L,
structure(list(Company = structure(c(2L, 2L, 2L, 2L, 2L, 2L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 3L, 3L), .Label = c("ABC Inc",
"ACME", "Handy Co"), class = "factor"), Quarter = structure(c(2L,
3L, 4L, 5L, 6L, 1L, 2L, 3L, 2L, 3L, 4L, 5L, 6L, 1L, 3L, 4L, 5L,
6L), .Label = c("1Q 2013", "1Q 2014", "1Q 2015", "2Q 2014", "3Q 2014",
"4Q 2014"), class = "factor"), Revenue = c(5000L, 6000L, 3200L,
1200L, 7200L, 10000L, 2500L, 4100L, 1250L, 8100L, 2300L, 3700L,
1100L, 1600L, 8000L, 9000L, 10000L, 12000L)), .Names = c("Company",
"Quarter", "Revenue"), class = "data.frame", row.names = c(NA,
-18L))
我希望将季度列重新分级为按年度排序，然后按季度排序，例如2013年第1季度、2014年第1季度、2014年第2季度、2015年第1季度。假设这只是一个更大数据集的样本，涉及几十年前的季度。我认为应该将季度分为两个字段：季度和年度。以下是使用dplyr的解决方案： 如果使用tidyr:：separate，您可以得到更简洁的解决方案：
我认为应该将季度分为两个字段：季度和年度。以下是使用dplyr的解决方案： 如果使用tidyr:：separate，您可以得到更简洁的解决方案：
这里有两种使用base r的方法
dat <- structure(list(Company = structure(c(2L, 2L, 2L, 2L, 2L, 2L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 3L, 3L), .Label = c("ABC Inc",
"ACME", "Handy Co"), class = "factor"), Quarter = structure(c(2L,
3L, 4L, 5L, 6L, 1L, 2L, 3L, 2L, 3L, 4L, 5L, 6L, 1L, 3L, 4L, 5L,
6L), .Label = c("1Q 2013", "1Q 2014", "1Q 2015", "2Q 2014", "3Q 2014",
"4Q 2014"), class = "factor"), Revenue = c(5000L, 6000L, 3200L,
1200L, 7200L, 10000L, 2500L, 4100L, 1250L, 8100L, 2300L, 3700L,
1100L, 1600L, 8000L, 9000L, 10000L, 12000L)), .Names = c("Company",
"Quarter", "Revenue"), class = "data.frame", row.names = c(NA,
-18L))
(tmp <- data.frame(do.call('rbind', strsplit(as.character(dat$Quarter), ' ')),
stringsAsFactors = FALSE))
# X1 X2
# 1 1Q 2014
# 2 1Q 2015
# 3 2Q 2014
# 4 3Q 2014
# 5 4Q 2014
# 6 1Q 2013
# 7 1Q 2014
# 8 1Q 2015
# 9 1Q 2014
# 10 1Q 2015
# 11 2Q 2014
# 12 3Q 2014
# 13 4Q 2014
# 14 1Q 2013
# 15 1Q 2015
# 16 2Q 2014
# 17 3Q 2014
# 18 4Q 2014
dat[order(tmp[, 2], tmp[, 1]), ]
# Company Quarter Revenue
# 6 ACME 1Q 2013 10000
# 14 ABC Inc 1Q 2013 1600
# 1 ACME 1Q 2014 5000
# 7 ABC Inc 1Q 2014 2500
# 9 ABC Inc 1Q 2014 1250
# 3 ACME 2Q 2014 3200
# 11 ABC Inc 2Q 2014 2300
# 16 Handy Co 2Q 2014 9000
# 4 ACME 3Q 2014 1200
# 12 ABC Inc 3Q 2014 3700
# 17 Handy Co 3Q 2014 10000
# 5 ACME 4Q 2014 7200
# 13 ABC Inc 4Q 2014 1100
# 18 Handy Co 4Q 2014 12000
# 2 ACME 1Q 2015 6000
# 8 ABC Inc 1Q 2015 4100
# 10 ABC Inc 1Q 2015 8100
# 15 Handy Co 1Q 2015 8000
这里有两种使用base r的方法
dat <- structure(list(Company = structure(c(2L, 2L, 2L, 2L, 2L, 2L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 3L, 3L), .Label = c("ABC Inc",
"ACME", "Handy Co"), class = "factor"), Quarter = structure(c(2L,
3L, 4L, 5L, 6L, 1L, 2L, 3L, 2L, 3L, 4L, 5L, 6L, 1L, 3L, 4L, 5L,
6L), .Label = c("1Q 2013", "1Q 2014", "1Q 2015", "2Q 2014", "3Q 2014",
"4Q 2014"), class = "factor"), Revenue = c(5000L, 6000L, 3200L,
1200L, 7200L, 10000L, 2500L, 4100L, 1250L, 8100L, 2300L, 3700L,
1100L, 1600L, 8000L, 9000L, 10000L, 12000L)), .Names = c("Company",
"Quarter", "Revenue"), class = "data.frame", row.names = c(NA,
-18L))
(tmp <- data.frame(do.call('rbind', strsplit(as.character(dat$Quarter), ' ')),
stringsAsFactors = FALSE))
# X1 X2
# 1 1Q 2014
# 2 1Q 2015
# 3 2Q 2014
# 4 3Q 2014
# 5 4Q 2014
# 6 1Q 2013
# 7 1Q 2014
# 8 1Q 2015
# 9 1Q 2014
# 10 1Q 2015
# 11 2Q 2014
# 12 3Q 2014
# 13 4Q 2014
# 14 1Q 2013
# 15 1Q 2015
# 16 2Q 2014
# 17 3Q 2014
# 18 4Q 2014
dat[order(tmp[, 2], tmp[, 1]), ]
# Company Quarter Revenue
# 6 ACME 1Q 2013 10000
# 14 ABC Inc 1Q 2013 1600
# 1 ACME 1Q 2014 5000
# 7 ABC Inc 1Q 2014 2500
# 9 ABC Inc 1Q 2014 1250
# 3 ACME 2Q 2014 3200
# 11 ABC Inc 2Q 2014 2300
# 16 Handy Co 2Q 2014 9000
# 4 ACME 3Q 2014 1200
# 12 ABC Inc 3Q 2014 3700
# 17 Handy Co 3Q 2014 10000
# 5 ACME 4Q 2014 7200
# 13 ABC Inc 4Q 2014 1100
# 18 Handy Co 4Q 2014 12000
# 2 ACME 1Q 2015 6000
# 8 ABC Inc 1Q 2015 4100
# 10 ABC Inc 1Q 2015 8100
# 15 Handy Co 1Q 2015 8000
欣赏细节！欣赏细节！我通常会同意把它们分开。不幸的是，在我的特殊情况下，我最终需要在我的代码中重新组合它们。我通常同意将它们分离出来。不幸的是，在我的特定情况下，我最终需要在以后的代码中重新组合它们。
dat[order(y, x), ]
# Company Quarter Revenue
# 6 ACME 1Q 2013 10000
# 14 ABC Inc 1Q 2013 1600
# 1 ACME 1Q 2014 5000
# 7 ABC Inc 1Q 2014 2500
# 9 ABC Inc 1Q 2014 1250
# 3 ACME 2Q 2014 3200
# 11 ABC Inc 2Q 2014 2300
# 16 Handy Co 2Q 2014 9000
# 4 ACME 3Q 2014 1200
# 12 ABC Inc 3Q 2014 3700
# 17 Handy Co 3Q 2014 10000
# 5 ACME 4Q 2014 7200
# 13 ABC Inc 4Q 2014 1100
# 18 Handy Co 4Q 2014 12000
# 2 ACME 1Q 2015 6000
# 8 ABC Inc 1Q 2015 4100
# 10 ABC Inc 1Q 2015 8100
# 15 Handy Co 1Q 2015 8000
(tmp <- data.frame(do.call('rbind', strsplit(as.character(dat$Quarter), ' ')),
stringsAsFactors = FALSE))
# X1 X2
# 1 1Q 2014
# 2 1Q 2015
# 3 2Q 2014
# 4 3Q 2014
# 5 4Q 2014
# 6 1Q 2013
# 7 1Q 2014
# 8 1Q 2015
# 9 1Q 2014
# 10 1Q 2015
# 11 2Q 2014
# 12 3Q 2014
# 13 4Q 2014
# 14 1Q 2013
# 15 1Q 2015
# 16 2Q 2014
# 17 3Q 2014
# 18 4Q 2014
dat[order(tmp[, 2], tmp[, 1]), ]
# Company Quarter Revenue
# 6 ACME 1Q 2013 10000
# 14 ABC Inc 1Q 2013 1600
# 1 ACME 1Q 2014 5000
# 7 ABC Inc 1Q 2014 2500
# 9 ABC Inc 1Q 2014 1250
# 3 ACME 2Q 2014 3200
# 11 ABC Inc 2Q 2014 2300
# 16 Handy Co 2Q 2014 9000
# 4 ACME 3Q 2014 1200
# 12 ABC Inc 3Q 2014 3700
# 17 Handy Co 3Q 2014 10000
# 5 ACME 4Q 2014 7200
# 13 ABC Inc 4Q 2014 1100
# 18 Handy Co 4Q 2014 12000
# 2 ACME 1Q 2015 6000
# 8 ABC Inc 1Q 2015 4100
# 10 ABC Inc 1Q 2015 8100
# 15 Handy Co 1Q 2015 8000