r：头数据取整 [英] r: rounding when head data
问题描述
我不知道如何获取r才能正确显示数据的开头。这是Swirl的任务，因此我必须弄清楚如何以Swirl会接受的方式编写代码。
Swirl希望最终的打印输出看起来像这样：
I am not able to figure out how to get r to correctly show the head of my data. This is an assignment for Swirl, so I have to figure out how to write my code in a way that Swirl will accept as the answer.
Swirl wants the final printout to look exactly like this:
## Pclass agecat Sex N survivors perc_survived
## <int> <fctr> <chr> <int> <int> <dbl>
## 1 Under 15 female 2 1 50.000000
## 1 Under 15 male 3 3 100.000000
## 1 15 to 50 female 70 68 97.142857
## 1 15 to 50 male 72 32 44.444444
## 1 Over 50 female 13 13 100.000000
## 1 Over 50 male 26 5 19.230769
#
我的代码：
library(dplyr)
titanic_4 <- titanic %>%
select(Survived, Pclass, Age, Sex) %>%
filter(!is.na(Age)) %>%
mutate(agecat = cut(Age, breaks = c(0, 14.99, 50, 150),
include.lowest = TRUE,
labels = c("Under 15", "15 to 50",
"Over 50"))) %>%
group_by(Pclass,agecat,Sex) %>%
summarize(N=n(), survivors = sum(Survived))%>%
mutate(perc_survived = (signif((100*survivors/N), digits=8)))
print(titanic_4)
礼物：
# A tibble: 18 x 6
# Groups: Pclass, agecat [9]
Pclass agecat Sex N survivors perc_survived
<int> <fctr> <chr> <int> <int> <dbl>
1 1 Under 15 female 2 1 50.000000
2 1 Under 15 male 3 3 100.000000
3 1 15 to 50 female 70 68 97.142857
4 1 15 to 50 male 72 32 44.444444
5 1 Over 50 female 13 13 100.000000
6 1 Over 50 male 26 5 19.230769
7 2 Under 15 female 10 10 100.000000
8 2 Under 15 male 9 9 100.000000
9 2 15 to 50 female 61 56 91.803279
10 2 15 to 50 male 78 5 6.410256
11 2 Over 50 female 3 2 66.666667
12 2 Over 50 male 12 1 8.333333
13 3 Under 15 female 27 13 48.148148
14 3 Under 15 male 27 9 33.333333
15 3 15 to 50 female 74 33 44.594595
16 3 15 to 50 male 217 29 13.364055
17 3 Over 50 female 1 1 100.000000
18 3 Over 50 male 9 0 0.000000
我转头时（titanic_4 ），r将最后一列中的数据四舍五入（perc_survivied）：
When I head(titanic_4), r rounds the data in the last column (perc_survivied):
# A tibble: 6 x 6
# Groups: Pclass, agecat [3]
Pclass agecat Sex N survivors perc_survived
<int> <fctr> <chr> <int> <int> <dbl>
1 1 Under 15 female 2 1 50.00000
2 1 Under 15 male 3 3 100.00000
3 1 15 to 50 female 70 68 97.14286
4 1 15 to 50 male 72 32 44.44444
5 1 Over 50 female 13 13 100.00000
6 1 Over 50 male 26 5 19.23077
但是，我希望R在perc_survived中给我小数点后六位，以便它看起来像这样：
However, I would like R to give me six decimal places in perc_survived so that it will look like this:
## Pclass agecat Sex N survivors perc_survived
## <int> <fctr> <chr> <int> <int> <dbl>
## 1 Under 15 female 2 1 50.000000
## 1 Under 15 male 3 3 100.000000
## 1 15 to 50 female 70 68 97.142857
## 1 15 to 50 male 72 32 44.444444
## 1 Over 50 female 13 13 100.000000
## 1 Over 50 male 26 5 19.230769
有人可以告诉我如何告诉r保留小数点后6位吗？
非常感谢！
Can anyone tell me how to tell r to keep 6 decimal place? Thank you so much!
来自评论：
*也许是print（titanic [1：6，]）？ – Florian
From Comments:
*Maybe print(titanic[1:6,])? – Florian
我尝试了Florian提出的方法，但并没有改变舍入结果
I tried the method proposed by Florian, but it did not change the rounding results
> titanic_4 <- titanic %>%
+ select(Survived, Pclass, Age, Sex) %>%
+ filter(!is.na(Age)) %>%
+ mutate(agecat = cut(Age, breaks = c(0, 14.99, 50, 150),
+ include.lowest = TRUE,
+ labels = c("Under 15", "15 to 50",
+ "Over 50"))) %>%
+ group_by(Pclass,agecat,Sex) %>%
+ summarize(N=n(), survivors = sum(Survived))%>%
+ mutate(perc_survived = (signif((100*survivors/N), digits=8)))
>
> print(titanic_4[1:6,])
# A tibble: 6 x 6
# Groups: Pclass, agecat [3]
Pclass agecat Sex N survivors perc_survived
<int> <fctr> <chr> <int> <int> <dbl>
1 1 Under 15 female 2 1 50.00000
2 1 Under 15 male 3 3 100.00000
3 1 15 to 50 female 70 68 97.14286
4 1 15 to 50 male 72 32 44.44444
5 1 Over 50 female 13 13 100.00000
6 1 Over 50 male 26 5 19.23077
>
关于Eric Fail的回答，sprintf导致列更改为字符。
With regards to the Answer by Eric Fail, sprintf causes the column to change to character. This is an assignment for Swirl(), and swirl will not accept the type change.
> titanic_4 <- titanic %>%
+ select(Survived, Pclass, Age, Sex) %>%
+ filter(!is.na(Age)) %>%
+ mutate(agecat = cut(Age, breaks = c(0, 14.99, 50, 150),
+ include.lowest = TRUE,
+ labels = c("Under 15", "15 to 50",
+ "Over 50"))) %>%
+ group_by(Pclass,agecat,Sex) %>%
+ summarize(N=n(), survivors = sum(Survived))%>%
+ mutate(perc_survived = sprintf("%0.6f",(signif((100*survivors/N), digits=8))))
>
> head (titanic_4)
# A tibble: 6 x 6
# Groups: Pclass, agecat [3]
Pclass agecat Sex N survivors perc_survived
<int> <fctr> <chr> <int> <int> <chr>
1 1 Under 15 female 2 1 50.000000
2 1 Under 15 male 3 3 100.000000
3 1 15 to 50 female 70 68 97.142857
4 1 15 to 50 male 72 32 44.444444
5 1 Over 50 female 13 13 100.000000
6 1 Over 50 male 26 5 19.230769
使用选项（位数= 8）的建议成功。为了使该建议生效，在运行我的代码之前，我必须更改r的基本选项，以便将其舍入到正确的位数。我的r设置为四舍五入。
The suggestion to use option(digits=8) was successful. In order to get this suggestion to work, before running my code, I had to change the basic options of r so that it would round to the right number of digits. My r was set to round to 5.
> options(digits=8)
> titanic_4 <- titanic %>%
+ select(Survived, Pclass, Age, Sex) %>%
+ filter(!is.na(Age)) %>%
+ mutate(agecat = cut(Age, breaks = c(0, 14.99, 50, 150),
+ include.lowest = TRUE,
+ labels = c("Under 15", "15 to 50",
+ "Over 50"))) %>%
+ group_by(Pclass,agecat,Sex) %>%
+ summarize(N=n(), survivors = sum(Survived))%>%
+ mutate(perc_survived = (round((100*survivors/N),digits=6)))
>
> head (titanic_4)
# A tibble: 6 x 6
# Groups: Pclass, agecat [3]
Pclass agecat Sex N survivors perc_survived
<int> <fctr> <chr> <int> <int> <dbl>
1 1 Under 15 female 2 1 50.000000
2 1 Under 15 male 3 3 100.000000
3 1 15 to 50 female 70 68 97.142857
4 1 15 to 50 male 72 32 44.444444
5 1 Over 50 female 13 13 100.000000
6 1 Over 50 male 26 5 19.230769
非常感谢您的评论和回答。
最好的祝福，
Thank you very much for your comments and answers. Best wishes,
提请
推荐答案
sprintf(c(.8693683839, .7869698963), fmt='%#.6g')
#> [1] "0.869368" "0.786970"
，特别是针对您的情况，
and specifically for your case,
titanic_4 <- tibble(perc_survived = c(50.000000, 100.000000, 97.142857,
44.444444, 100.000000, 19.230769))
titanic_4
#> # A tibble: 6 x 1
#> perc_survived
#> <dbl>
#> 1 50.00000
#> 2 100.00000
#> 3 97.14286
#> 4 44.44444
#> 5 100.00000
#> 6 19.23077
#> >
>
titanic_4 <- titanic_4 %>% mutate(perc_survived_6 = sprintf("%0.6f", perc_survived))
titanic_4
#> # A tibble: 6 x 2
#> perc_survived perc_survived_6
#> <dbl> <chr>
#> 1 50.00000 50.000000
#> 2 100.00000 100.000000
#> 3 97.14286 97.142857
#> 4 44.44444 44.444444
#> 5 100.00000 100.000000
#> 6 19.23077 19.230769
或可能更改全局位数，
options(digits=8)
titanic_4
#> # A tibble: 6 x 1
#> perc_survived
#> <dbl>
#> 1 50.000000
#> 2 100.000000
#> 3 97.142857
#> 4 44.444444
#> 5 100.000000
#> 6 19.230769
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