我有一张这样的表：
City; Class; Number; Sum
BE ; 01; 734; 4711
BE ; 02; 896; 4711
BE ; 03; 1258; 4711
BE ; 04; 980; 4711
BE ; 05; 543; 4711
BE ; 06; 192; 4711
BE ; 07; 69; 4711
BE ; 08; 20; 4711
BE ; 09; 14; 4711
BE ; 10; 4; 4711
BE ; 12; 1; 4711
FR ; 01; 1213;14258
FR ; 02; 2217;14258
FR ; 03; 3369;14258
FR ; 04; 4037;14258
FR ; 05; 2117;14258
FR ; 06; 774;14258
FR ; 07; 301;14258
FR ; 08; 124;14258
FR ; 09; 62;14258
FR ; 10; 21;14258
FR ; 11; 11;14258
FR ; 12; 4;14258
FR ; 13; 2;14258
FR ; 14; 3;14258
FR ; 16; 3;14258
我想在“Class”变量的8个范围内“分组”：
范围1 =“01”，范围2 =“02”，范围3 =“03”，范围4 =“04”，范围5 =“05”，以及3个范围的几个值：范围6 =“06 “，”07，“08”，“09，”10“， 范围7 =“11”，“12”，“13”，“14”，“15”， 范围8 =“16”，“17”，“18”，“19”，“20”。
是否可以使用group_by执行此操作？
变量“City”和“Class”是字符格式。
非常感谢您的帮助。
输出表应如下所示：
City; Range; Number; Sum;
BE ; R1; 734; 4711;
BE ; R2; 896; 4711;
BE ; R3; 1258; 4711;
BE ; R4; 980; 4711;
BE ; R5; 543; 4711;
BE ; R6; 299; 4711;
BE ; R7; 1; 4711;
FR ; R1; 1213;14258;
FR ; R2; 2217;14258;
FR ; R3; 3369;14258;
FR ; R4; 4037;14258;
FR ; R5; 2117;14258;
FR ; R6; 1282;14258;
FR ; R7; 20;14258;
FR ; R8; 3;14258;
非常感谢。
答案 0 :(得分：3)
使用cut，您可以根据breaks值创建范围，并使用它创建range列。然后通过对Number变量进行分组来计算City的总和。
df$Range <- cut( as.numeric(df$Class), breaks = c(0, 1,2,3,4,5,10,15,20 ), labels = paste('R', 1:8, sep = '' )) # create Range column
df$Class <- NULL # remove Class column
df <- within( df, Sum <- ave(Number, City, FUN = sum ) ) # compute sum by grouping City
df <- within(df, Number <- ave( Number, by = list(City, Range), FUN = sum ) ) # compute sum by grouping City and Range
df[ !duplicated(df), ] # remove duplicated rows
# City Number Sum Range
# 1 BE 734 4711 R1
# 2 BE 896 4711 R2
# 3 BE 1258 4711 R3
# 4 BE 980 4711 R4
# 5 BE 543 4711 R5
# 6 BE 299 4711 R6
# 11 BE 1 4711 R7
# 12 FR 1213 14258 R1
# 13 FR 2217 14258 R2
# 14 FR 3369 14258 R3
# 15 FR 4037 14258 R4
# 16 FR 2117 14258 R5
# 17 FR 1282 14258 R6
# 22 FR 20 14258 R7
# 26 FR 3 14258 R8
数据：
df <- structure(list(City = c("BE", "BE", "BE", "BE", "BE", "BE", "BE",
"BE", "BE", "BE", "BE", "FR", "FR", "FR", "FR", "FR", "FR", "FR",
"FR", "FR", "FR", "FR", "FR", "FR", "FR", "FR"),
Class = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 12L, 1L, 2L, 3L, 4L, 5L,
6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 16L),
Number = c(734L, 896L, 1258L, 980L, 543L, 192L, 69L, 20L, 14L, 4L, 1L, 1213L,
2217L, 3369L, 4037L, 2117L, 774L, 301L, 124L, 62L, 21L, 11L,
4L, 2L, 3L, 3L),
Sum = c(4711L, 4711L, 4711L, 4711L, 4711L, 4711L, 4711L, 4711L, 4711L, 4711L,
4711L, 14258L, 14258L, 14258L, 14258L, 14258L, 14258L, 14258L, 14258L,
14258L, 14258L, 14258L, 14258L, 14258L, 14258L, 14258L)),
.Names = c("City", "Class", "Number", "Sum"),
row.names = c(NA, -26L), class = "data.frame")
答案 1 :(得分：2)
我们使用原始数据集创建key/value数据集join，按“类”分组并获取“数字”的sum
library(dplyr)
keyDat <- data.frame(Class = sprintf("%02d", 1:20),
range = rep(paste0("", 1:8), rep(c(1, 5), c(5, 3))), stringsAsFactors=FALSE)
df1 %>%
left_join(., keyDat) %>%
group_by(City, Range = range) %>%
summarise(Number = sum(Number), Sum= Sum[1L])
# City Range Number Sum
# <chr> <chr> <int> <int>
#1 BE R1 734 4711
#2 BE R2 896 4711
#3 BE R3 1258 4711
#4 BE R4 980 4711
#5 BE R5 543 4711
#6 BE R6 299 4711
#7 BE R7 1 4711
#8 FR R1 1213 14258
#9 FR R2 2217 14258
#10 FR R3 3369 14258
#11 FR R4 4037 14258
#12 FR R5 2117 14258
#13 FR R6 1282 14258
#14 FR R7 20 14258
#15 FR R8 3 14258
df1 <- structure(list(City = c("BE ", "BE ", "BE ", "BE ", "BE ", "BE ",
"BE ", "BE ", "BE ", "BE ", "BE ", "FR ", "FR ", "FR ", "FR ",
"FR ", "FR ", "FR ", "FR ", "FR ", "FR ", "FR ", "FR ", "FR ",
"FR ", "FR "), Class = c("01", "02", "03", "04", "05", "06",
"07", "08", "09", "10", "12", "01", "02", "03", "04", "05", "06",
"07", "08", "09", "10", "11", "12", "13", "14", "16"), Number = c(734L,
896L, 1258L, 980L, 543L, 192L, 69L, 20L, 14L, 4L, 1L, 1213L,
2217L, 3369L, 4037L, 2117L, 774L, 301L, 124L, 62L, 21L, 11L,
4L, 2L, 3L, 3L), Sum = c(4711L, 4711L, 4711L, 4711L, 4711L, 4711L,
4711L, 4711L, 4711L, 4711L, 4711L, 14258L, 14258L, 14258L, 14258L,
14258L, 14258L, 14258L, 14258L, 14258L, 14258L, 14258L, 14258L,
14258L, 14258L, 14258L)), .Names = c("City", "Class", "Number",
"Sum"), row.names = c(NA, -26L), class = "data.frame")
答案 2 :(得分：1)
Before updating someNum is 2.
After updating someNum is 3.
In main, someNum is now 3.
Before updating someNum is 3.
After updating someNum is 4.
In main, someNum is now 4.