dplyr代码工作,现在不是,我不知道为什么
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我在一周前编写了一些代码,这些代码效果很好,但现在它已经停止工作了。我已经更新了tidyverse包,看看这可能是问题,但事实并非如此。
以下是一些示例数据:
yearmo sex eco ue fs12ago fs12ahead purchases
200301 Male neutral negative negative neutral neutral
200301 Female negative negative negative neutral neutral
200301 Female negative negative neutral neutral positive
200301 Male neutral neutral neutral neutral neutral
200301 Male negative negative negative positive negative
200301 Male negative negative neutral neutral positive
200301 Male negative negative neutral neutral neutral
200301 Male negative negative positive neutral negative
200301 Female negative negative negative neutral positive
200301 Female negative negative positive negative neutral
200301 Female negative negative negative negative negative
200301 Female negative neutral negative neutral negative
200301 Male negative negative neutral neutral negative
200301 Female positive neutral neutral neutral positive
200301 Male negative negative neutral neutral positive
200301 Male neutral negative negative neutral neutral
200301 Female neutral negative negative neutral neutral
200301 Male neutral negative neutral neutral positive
200301 Female negative negative negative negative positive
200301 Female positive negative neutral neutral positive
正在运行的代码,但现在不是,这是:
tmp_eco <- data %>%
group_by(yearmo) %>%
count(yearmo, eco)
我以前获得的输出是对于称为“eco”的变量响应为正,负或中性的人数,例如:
yearmo eco n
200301 positive 10
200301 negative 13
200301 neutral 9
200301 positive 7
200301 negative 5
200301 neutral 16
我现在得到的错误是:
Error: Can't subset with `[` using an object of class quoted.
Call `rlang::last_error()` to see a backtrace
这给了我:
<error>
message: Can't subset with `[` using an object of class quoted.
class: `rlang_error`
backtrace:
1. dplyr::group_by(., yearmo)
9. plyr::count(., yearmo, eco)
14. plyr::eval.quoted(vars, df)
18. tibble:::`[.tbl_df`(envir, exprs)
19. tibble:::check_names_df(i, x)
Call `rlang::last_trace()` to see the full backtrace
有关为什么会发生这种情况的任何想法?
答案
随着'数据'显示,如果我们使用group_by
,则不需要count
library(dplyr)
data %>%
dplyr::count(yearmo, eco)
# A tibble: 3 x 3
# yearmo eco n
# <int> <chr> <int>
#1 200301 negative 13
#2 200301 neutral 5
#3 200301 positive 2
data
data <- structure(list(yearmo = c(200301L, 200301L, 200301L, 200301L,
200301L, 200301L, 200301L, 200301L, 200301L, 200301L, 200301L,
200301L, 200301L, 200301L, 200301L, 200301L, 200301L, 200301L,
200301L, 200301L), sex = c("Male", "Female", "Female", "Male",
"Male", "Male", "Male", "Male", "Female", "Female", "Female",
"Female", "Male", "Female", "Male", "Male", "Female", "Male",
"Female", "Female"), eco = c("neutral", "negative", "negative",
"neutral", "negative", "negative", "negative", "negative", "negative",
"negative", "negative", "negative", "negative", "positive", "negative",
"neutral", "neutral", "neutral", "negative", "positive"), ue = c("negative",
"negative", "negative", "neutral", "negative", "negative", "negative",
"negative", "negative", "negative", "negative", "neutral", "negative",
"neutral", "negative", "negative", "negative", "negative", "negative",
"negative"), fs12ago = c("negative", "negative", "neutral", "neutral",
"negative", "neutral", "neutral", "positive", "negative", "positive",
"negative", "negative", "neutral", "neutral", "neutral", "negative",
"negative", "neutral", "negative", "neutral"), fs12ahead = c("neutral",
"neutral", "neutral", "neutral", "positive", "neutral", "neutral",
"neutral", "neutral", "negative", "negative", "neutral", "neutral",
"neutral", "neutral", "neutral", "neutral", "neutral", "negative",
"neutral"), purchases = c("neutral", "neutral", "positive", "neutral",
"negative", "positive", "neutral", "negative", "positive", "neutral",
"negative", "negative", "negative", "positive", "positive", "neutral",
"neutral", "positive", "positive", "positive")),
class = "data.frame", row.names = c(NA,
-20L))
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