我正在尝试运行R: extract maximum value in vector under certain conditions中的代码,但我一直收到错误
Error in list(id.2 = c(3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, :
invalid subscript type 'integer'
代码如下:
require(dplyr)
dat <- read.table(header = TRUE, text = "id name year job job2 cumu_job2
1 Jane 1980 Worker 0 0
1 Jane 1981 Manager 1 1
1 Jane 1982 Sales 0 0
1 Jane 1983 Sales 0 0
1 Jane 1984 Manager 1 1
1 Jane 1985 Manager 1 2
1 Jane 1986 Boss 0 0
2 Bob 1985 Worker 0 0
2 Bob 1986 Sales 0 0
2 Bob 1987 Manager 1 1
2 Bob 1988 Manager 1 2
2 Bob 1989 Boss 0 0
3 Jill 1989 Worker 0 0
3 Jill 1990 Boss 0 0")
dat %.%
group_by(id) %.%
mutate(
all_jobs = sum(unique(job) %in% c("Sales","Manager","Boss")),
cumu_max = max(cumu_job2)
) %.%
filter(all_jobs == 3, job %in% c("Sales","Boss"))
Source: local data frame [5 x 8]
Groups: id
id name year job job2 cumu_job2 all_jobs cumu_max
1 1 Jane 1982 Sales 0 0 3 2
2 1 Jane 1983 Sales 0 0 3 2
3 1 Jane 1986 Boss 0 0 3 2
4 2 Bob 1986 Sales 0 0 3 2
5 2 Bob 1989 Boss 0 0 3 2
答案 0 :(得分:10)
示例代码也适用于我。但我发现如果我尝试这样做,我可以重现类似的错误:
dat %.%
group_by(dat$id) %.%
mutate(
all_jobs = sum(unique(job) %in% c("Sales","Manager","Boss")),
cumu_max = max(cumu_job2)
) %.%
filter(all_jobs == 3, job %in% c("Sales","Boss"))
也就是说,如果我输入“group_by(dat $ id)”而不是“group_by(id)”
答案 1 :(得分:6)
<强>错误强>
示例代码也适用于我。但是,正如schnee所提到的,您可以通过group_by(dat $ id)替换group_by(id)来创建类似的错误。可重现的代码:
dat1[,'x']
虽然第一个通常只是一个拼写错误(你可以用x替换dat $ x),但第二个可能是一个有效的用例(尽管我建议使用连接来使其更清晰)。
<强>解决方案强>
dplyr包不喜欢&#39; $&#39;的使用。请尝试使用&#39; [&#39;,例如:
dat1$'x'
引用变量也有效:
dat1 %>%
group_by(dat1[,'x']) %>%
mutate(y = sum(unique(y) %in% c("A","B","C")))
dat1 %>%
group_by(dat1$'x') %>%
mutate(y = sum(unique(y) %in% c("A","B","C")))
完整代码:
<script type="text/ng-template" id="template1.html">
<div class="modal_content"> {{data[0]}}</div>
</script>
另请参阅http://goo.gl/2hdIra或https://github.com/hadley/dplyr/issues/433