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我正在处理来自 r Programming Environment 12 Data Manipulation 的一个问题。我不知道如何让 r 在小数点后给我正确的位数。

我的代码:

 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 = (survivors/N)*100.000000)

head(titanic_4)

给出:

# 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 中给我六个小数位,以便它看起来像这样:

## 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

谁能告诉我如何告诉 r 保留小数点后 6 位?

我试过 sprintf:

> 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("%.6f",((survivors/N)*100.000000)))
> 
> head(titanic_4)

这使:

# 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

添加 sprintf 可以纠正小数位的问题,但它产生了一个新问题。sprintf 将列类型从 更改<dbl><chr>

漩涡不会接受这个答案。有谁知道另一种方式?

太感谢了!

4

2 回答 2

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sprintf是一个字符串操作函数,所以根据定义它会返回一个字符串。如果您只是想四舍五入到一组数字,要么roundsignif(重要数字)应该可以工作。两者都有要保留的位数的参数。所以它似乎mutate(perc_survived = round((survivors / N) * 100, digits = 6))会给你你正在寻找的东西。如果您想要有效数字的数量,而不是简单的四舍五入,请使用signif.

于 2018-01-25T16:39:08.177 回答
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您可以使用

> sprintf("%.6f", .1)
[1] "0.100000"
于 2018-01-19T14:03:53.187 回答