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我有以下数据

df <- structure(list(year = c(2015L, 2015L, 2015L, 2015L, 2015L, 2015L, 
2015L, 2015L, 2015L, 2015L, 2015L, 2015L, 2015L, 2015L, 2015L, 
2015L, 2015L, 2015L, 2015L, 2015L, 2015L, 2015L, 2015L, 2015L, 
2015L, 2015L, 2016L, 2016L, 2016L, 2016L, 2016L, 2016L, 2016L, 
2016L, 2016L, 2016L, 2016L, 2016L, 2016L, 2016L, 2016L, 2016L, 
2016L, 2016L, 2016L, 2016L, 2016L, 2016L, 2016L, 2016L, 2016L, 
2016L), newly_engaged = c(FALSE, FALSE, FALSE, FALSE, FALSE, 
FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, TRUE, 
TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, 
TRUE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, 
FALSE, FALSE, FALSE, FALSE, FALSE, TRUE, TRUE, TRUE, TRUE, TRUE, 
TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE), qualification = structure(c(1L, 
1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 
2L, 2L, 2L), .Label = c("A2", "AS"), class = "factor"), subject = structure(c(7L, 
7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 
7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 
7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 
7L, 7L, 7L), .Label = c("Biology", "Chemistry", "Mathematics", 
"Mathematics (Further)", "Mathematics (Pure)", "Mathematics (Statistics)", 
"Physics"), class = "factor"), grade = structure(c(1L, 2L, 3L, 
4L, 5L, 6L, 7L, 2L, 3L, 4L, 5L, 6L, 7L, 1L, 2L, 3L, 4L, 5L, 6L, 
7L, 2L, 3L, 4L, 5L, 6L, 7L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 2L, 3L, 
4L, 5L, 6L, 7L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 2L, 3L, 4L, 5L, 6L, 
7L), .Label = c("S", "A", "B", "C", "D", "E", "No.results"), class = "factor"), 
    c = c(2032L, 3871L, 3728L, 3130L, 2514L, 1796L, 591L, 7694L, 
    5486L, 4885L, 3790L, 2493L, 2734L, 1079L, 2142L, 2082L, 1703L, 
    1273L, 779L, 219L, 4096L, 2880L, 2366L, 1700L, 1139L, 1051L, 
    1807L, 3961L, 3921L, 3237L, 2521L, 1760L, 609L, 8160L, 6661L, 
    7035L, 5934L, 4811L, 6155L, 1009L, 2022L, 2127L, 1664L, 1224L, 
    779L, 192L, 4214L, 3350L, 3336L, 2701L, 2044L, 2280L), e = c(17662L, 
    17662L, 17662L, 17662L, 17662L, 17662L, 17662L, 27082L, 27082L, 
    27082L, 27082L, 27082L, 27082L, 9277L, 9277L, 9277L, 9277L, 
    9277L, 9277L, 9277L, 13232L, 13232L, 13232L, 13232L, 13232L, 
    13232L, 17816L, 17816L, 17816L, 17816L, 17816L, 17816L, 17816L, 
    38756L, 38756L, 38756L, 38756L, 38756L, 38756L, 9017L, 9017L, 
    9017L, 9017L, 9017L, 9017L, 9017L, 17925L, 17925L, 17925L, 
    17925L, 17925L, 17925L), m = c(0.115049258294644, 0.219171101800476, 
    0.211074623485449, 0.177216623258974, 0.142339485901936, 
    0.101687238138376, 0.0334616691201449, 0.2841001403146, 0.202569972675578, 
    0.180378110922384, 0.139945351155749, 0.0920537626467765, 
    0.100952662284912, 0.116309151665409, 0.230893607847364, 
    0.224425999784413, 0.183572275520103, 0.137221084402285, 
    0.0839711113506521, 0.0236067694297726, 0.309552599758162, 
    0.217654171704958, 0.178808948004837, 0.128476420798065, 
    0.0860792019347038, 0.0794286577992745, 0.101425684777728, 
    0.222328244274809, 0.220083071396498, 0.181690615177369, 
    0.14150202065559, 0.0987876066457117, 0.0341827570722946, 
    0.210548044173805, 0.171870162039426, 0.181520280730726, 
    0.153111776241098, 0.124135617710806, 0.158814119104139, 
    0.11189974492625, 0.224243096373517, 0.235887767550183, 0.184540312742597, 
    0.135743595430853, 0.0863923699678385, 0.0212931130087612, 
    0.235090655509066, 0.186889818688982, 0.186108786610879, 
    0.15068340306834, 0.114030683403068, 0.127196652719665)), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -52L), .Names = c("year", 
"newly_engaged", "qualification", "subject", "grade", "c", "e", 
"m"))

我需要取m2015 年和 2016 年对应值的差异，以显示 2015 年到 2016 年分配的成绩比例的差异。我想我可以reshape2::cast这样做并ddplyr::summarise计算差异，但我不确定如何首先使用cast。

4

2 回答 2

1

如果我们与plyr库一起加载库，则会发生错误，dplyr因为两者中的函数名称相同，并且这些函数可能会被其他包屏蔽

df %>%
   group_by(year) %>%
   plyr::mutate(n = row_number()) %>% 
   group_by(n) %>% 
   summarise(m = diff(m))

排名错误（x，ties.method = “first”，na.last = “keep”）：
缺少参数“x”，没有默认值

在这种情况下，dply::明确指定

df %>% 
   group_by(year) %>% 
   dplyr::mutate(n = row_number()) %>% 
   group_by(n) %>% 
   dplyr::summarise(m = diff(m)) 
# A tibble: 26 × 2
#      n             m
#   <int>         <dbl>
#1      1 -0.0136235735
#2      2  0.0031571425
#3      3  0.0090084479
#4      4  0.0044739919
#5      5 -0.0008374652
#6      6 -0.0028996315
#7      7  0.0007210880
#8      8 -0.0735520961
#9      9 -0.0306998106
#10    10  0.0011421698
# ... with 16 more rows
于 2017-04-20T15:58:38.953 回答
1

使用dplyr并且tidyr您可以轻松地重铸您的数据框以给出 2015 年和 2016 年的 m 值，然后计算差异

library(dplyr)
library(tidyr)
df2 <- df %>% select(-c(c,e)) %>% spread(key=year,value=m) %>% mutate(diff=`2016`-`2015`)

df2
# A tibble: 26 × 7
   newly_engaged qualification subject      grade     `2015`     `2016`          diff
           <lgl>        <fctr>  <fctr>     <fctr>      <dbl>      <dbl>         <dbl>
1          FALSE            A2 Physics          S 0.11504926 0.10142568 -0.0136235735
2          FALSE            A2 Physics          A 0.21917110 0.22232824  0.0031571425
3          FALSE            A2 Physics          B 0.21107462 0.22008307  0.0090084479
4          FALSE            A2 Physics          C 0.17721662 0.18169062  0.0044739919
5          FALSE            A2 Physics          D 0.14233949 0.14150202 -0.0008374652
6          FALSE            A2 Physics          E 0.10168724 0.09878761 -0.0028996315
7          FALSE            A2 Physics No.results 0.03346167 0.03418276  0.0007210880
8          FALSE            AS Physics          A 0.28410014 0.21054804 -0.0735520961
9          FALSE            AS Physics          B 0.20256997 0.17187016 -0.0306998106
10         FALSE            AS Physics          C 0.18037811 0.18152028  0.0011421698
# ... with 16 more rows
于 2017-04-20T15:59:41.373 回答