操作上基于可变

列我有以下数据操作上基于可变

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"))

和我需要采取的m的对应值的差为2015和2016，以显示等级的分配的比例的差从2015年到2016年。我想我可以reshape2::cast这个和ddplyr::summarise来计算差异，但我不知道如何使用cast摆在首位。

来源

2017-04-20 Morpheu5

难道你的意思是'DF $ M [df $ year == 2016] - df $ m [df $ year == 2015]'或与'dplyr'即ie'df％>％group_by（year）％>％mutate（n = row_number（））％>％ group_by（n）％>％summary（m = diff（m））' – akrun

我认为这就是我的意思，但是我得到了一个错误（x，tie.method =“first”，na.last =“keep “）：参数”x“丢失，”dplyr“soluti没有默认值 – Morpheu5

这个错误与你展示的例子有关。它与我工作正常使用dplyr 0.5.0 – akrun

使用dplyr和tidyr您可以轻松地改写你的数据帧给m的值，2015年和2016年对方一起，然后计算差值

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

来源

2017-04-20 15:59:41

如果我们加载沿plyr库会发生错误的dplyr，因为是在两个相同的函数名以及这些功能可以得到掩盖了其他包中排名

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

错误（X，ties.method = “第一”，na.last = “保持”）：
参数 “X” 的缺失，没有默认设置

在这种情况下，指定dply::前plicitly

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

来源

2017-04-20 15:58:38 akrun

我很确定我没有plyr加载，我仍然得到错误...怪异的。 – Morpheu5

@ Morpheu5你可以检查'sessionInfo（）'或者尝试我用'dplyr ::'建议的方法 – akrun

操作上基于可变

回答

相关问题