R 将数据帧的列传递给内部函数
我希望将列中的管道插入到一个函数中,该函数使用自定义内部函数执行R 将数据帧的列传递给内部函数,r,tidyeval,non-standard-evaluation,R,Tidyeval,Non Standard Evaluation,我希望将列中的管道插入到一个函数中,该函数使用自定义内部函数执行purr::imap_dfr 我的目标是df%>%diffmean(df,group,col1,col2)将运行t.test(col1~group,.data=df)和t.test(col2~group,.data=df) ttests <- function(df, group, ...) { group <- rlang::ensym(group) vars <- rlang::ensyms(...)
purr::imap_dfr
我的目标是df%>%diffmean(df,group,col1,col2)
将运行t.test(col1~group,.data=df)
和t.test(col2~group,.data=df
)
ttests <- function(df, group, ...) {
group <- rlang::ensym(group)
vars <- rlang::ensyms(...)
df %>%
dplyr::select(c(!!!vars)) %>%
purrr::imap_dfr(function(.x, .y) {
broom::tidy(t.test(.x ~ !!group)) %>%
dplyr::mutate(name = .y) %>%
dplyr::select(name, dplyr::everything())
})
}
产生:
name estimate estimate1 estimate2 statistic p.value parameter conf.low conf.high method alternative
<chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <chr> <chr>
1 depth -0.247 61.5 61.8 -4.86 0.00000143 808. -0.347 -0.147 Welch Two Sample t-test two.sided
2 table 0.263 57.7 57.5 3.13 0.00183 805. 0.0977 0.427 Welch Two Sample t-test two.sided
3 price -3477. 506. 3983. -197. 0 51886. -3512. -3443. Welch Two Sample t-test two.sided
4 x -1.77 3.99 5.76 -299. 0 6451. -1.78 -1.76 Welch Two Sample t-test two.sided
5 y -1.75 4.01 5.76 -290. 0 6529. -1.76 -1.73 Welch Two Sample t-test two.sided
6 z -1.10 2.46 3.55 -294. 0 6502. -1.10 -1.09 Welch Two Sample t-test two.sided
name estimate1 estimate2统计p.value参数conf.low conf.high方法备选方案
1深度-0.247 61.5 61.8-4.86 0.00000143 808。-0.347-0.147韦尔奇双面双样本t检验
表0.263 57.7 57.5 3.13 0.00183 805.0.0977 0.427韦尔奇双样本t检验双侧
3价格-3477.506.3983.-197.0 51886.-3512.-3443.韦尔奇双样本t检验双面
4X-1.773.995.76-299.06451-1.78-1.76韦尔奇双面双样本t检验
5Y-1.754.015.76-290.06529-1.76-1.73韦尔奇双样本t检验双侧
6 z-1.10 2.46 3.55-294.0 6502。-1.10-1.09韦尔奇双面双样本t检验
基本Rt.test
命令并非真正设计用于rlang
样式的styntax,因此您需要对公式进行一些修改。这应该可以工作
ttests <- function(df, group, ...) {
group <- rlang::ensym(group)
vars <- rlang::ensyms(...)
df %>%
dplyr::select(c(!!!vars)) %>%
purrr::imap_dfr(function(.x, .y) {
rlang::eval_tidy(rlang::quo(t.test(!!rlang::sym(.y) ~ !!group, df))) %>%
broom::tidy() %>%
dplyr::mutate(name = .y) %>%
dplyr::select(name, dplyr::everything())
})
}
基本的R
t.test
命令并不是真正设计用于rlang
样式的styntax,因此您需要对公式进行一些修改。这应该可以工作
ttests <- function(df, group, ...) {
group <- rlang::ensym(group)
vars <- rlang::ensyms(...)
df %>%
dplyr::select(c(!!!vars)) %>%
purrr::imap_dfr(function(.x, .y) {
rlang::eval_tidy(rlang::quo(t.test(!!rlang::sym(.y) ~ !!group, df))) %>%
broom::tidy() %>%
dplyr::mutate(name = .y) %>%
dplyr::select(name, dplyr::everything())
})
}
还可以选择转换为“长”格式,然后在执行
nest\u by
library(dplyr)
library(tidyr)
ttests <- function(df, group, ...) {
grp <- rlang::as_name(ensym(group))
df %>%
dplyr::select(!!! enquos(...), grp) %>%
pivot_longer(cols = -grp) %>%
nest_by(name) %>%
transmute(name,
new = list(broom::tidy(t.test(reformulate(grp, response = 'value'), data)))) %>%
unnest_wider(c(new))
}
ttests(diamonds, carat, depth, table, price, x, y, z)
# A tibble: 6 x 11
# name estimate estimate1 estimate2 statistic p.value parameter conf.low conf.high method alternative
# <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <chr> <chr>
#1 depth -0.247 61.5 61.8 -4.86 0.00000143 808. -0.347 -0.147 Welch Two Sample t-test two.sided
#2 price -3477. 506. 3983. -197. 0 51886. -3512. -3443. Welch Two Sample t-test two.sided
#3 table 0.263 57.7 57.5 3.13 0.00183 805. 0.0977 0.427 Welch Two Sample t-test two.sided
#4 x -1.77 3.99 5.76 -299. 0 6451. -1.78 -1.76 Welch Two Sample t-test two.sided
#5 y -1.75 4.01 5.76 -290. 0 6529. -1.76 -1.73 Welch Two Sample t-test two.sided
#6 z -1.10 2.46 3.55 -294. 0 6502. -1.10 -1.09 Welch Two Sample t-test two.sided
库(dplyr)
图书馆(tidyr)
ttests%
枢轴长度(cols=-grp)%>%
嵌套单位(名称)%>%
蜕变,
新建=列表(扫帚::整洁(t.test(重新格式化(grp,响应='value'),数据)))%>%
unnest_加宽(c(新))
}
测试(钻石、克拉、深度、表、价格、x、y、z)
#一个tibble:6x11
#名称估计估计1估计2统计p.value参数conf.low conf.high方法备选方案
#
#1深度-0.247 61.5 61.8-4.86 0.00000143 808。-0.347-0.147韦尔奇双面双样本t检验
#2价格-3477.506.3983.-197.0 51886.-3512.-3443.韦尔奇双样本t检验双面
#表0.263 57.7 57.5 3.13 0.00183 805.0.0977 0.427 Welch双面双样本t检验
#4X-1.773.995.76-299.06451-1.78-1.76韦尔奇双面双样本t检验
#5Y-1.754.015.76-290.06529-1.76-1.73韦尔奇双样本t检验双侧
#6 z-1.10 2.46 3.55-294.0 6502。-1.10-1.09韦尔奇双面双样本t检验
另一个选项是转换为“长”格式,然后在执行嵌套操作后应用公式
library(dplyr)
library(tidyr)
ttests <- function(df, group, ...) {
grp <- rlang::as_name(ensym(group))
df %>%
dplyr::select(!!! enquos(...), grp) %>%
pivot_longer(cols = -grp) %>%
nest_by(name) %>%
transmute(name,
new = list(broom::tidy(t.test(reformulate(grp, response = 'value'), data)))) %>%
unnest_wider(c(new))
}
ttests(diamonds, carat, depth, table, price, x, y, z)
# A tibble: 6 x 11
# name estimate estimate1 estimate2 statistic p.value parameter conf.low conf.high method alternative
# <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <chr> <chr>
#1 depth -0.247 61.5 61.8 -4.86 0.00000143 808. -0.347 -0.147 Welch Two Sample t-test two.sided
#2 price -3477. 506. 3983. -197. 0 51886. -3512. -3443. Welch Two Sample t-test two.sided
#3 table 0.263 57.7 57.5 3.13 0.00183 805. 0.0977 0.427 Welch Two Sample t-test two.sided
#4 x -1.77 3.99 5.76 -299. 0 6451. -1.78 -1.76 Welch Two Sample t-test two.sided
#5 y -1.75 4.01 5.76 -290. 0 6529. -1.76 -1.73 Welch Two Sample t-test two.sided
#6 z -1.10 2.46 3.55 -294. 0 6502. -1.10 -1.09 Welch Two Sample t-test two.sided
库(dplyr)
图书馆(tidyr)
ttests%
枢轴长度(cols=-grp)%>%
嵌套单位(名称)%>%
蜕变,
新建=列表(扫帚::整洁(t.test(重新格式化(grp,响应='value'),数据)))%>%
unnest_加宽(c(新))
}
测试(钻石、克拉、深度、表、价格、x、y、z)
#一个tibble:6x11
#名称估计估计1估计2统计p.value参数conf.low conf.high方法备选方案
#
#1深度-0.247 61.5 61.8-4.86 0.00000143 808。-0.347-0.147韦尔奇双面双样本t检验
#2价格-3477.506.3983.-197.0 51886.-3512.-3443.韦尔奇双样本t检验双面
#表0.263 57.7 57.5 3.13 0.00183 805.0.0977 0.427 Welch双面双样本t检验
#4X-1.773.995.76-299.06451-1.78-1.76韦尔奇双面双样本t检验
#5Y-1.754.015.76-290.06529-1.76-1.73韦尔奇双样本t检验双侧
#6 z-1.10 2.46 3.55-294.0 6502。-1.10-1.09韦尔奇双面双样本t检验
这太好了,谢谢!我还想检查group
是否是逻辑向量-rlang::eval_tidy(rlang::quo(is.logical(df$!!group)))
不起作用,因为它不喜欢df$!!group
。有什么简单的方法来评估它吗?抱歉,仍然在与NSE斗争-试图同时阅读Hadley在Advanced R中的章节。只需使用dplyr::pull()
动词:is.logical(df%>%pull(!!group))就更容易了
这太好了,谢谢!我还想检查group
是否是逻辑向量-rlang::eval_tidy(rlang::quo(is.logical(df$!!group)))
不起作用,因为它不喜欢df$!!group
。有什么简单的方法来评估这个问题吗?对不起,仍然在为NSE而挣扎-试图同时阅读Hadley在Advanced R中的章节。只需使用dplyr::pull()
动词:is.logical(df%>%pull(!!group))