Question

这是我的data_frame对象：

structure(list(dt = structure(c(17702, 17702, 17702, 17702, 17703, 
17703, 17704, 17705, 17705, 17706, 17706, 17706, 17706), class = "Date"), 
    uuid_lev = c(4L, 5L, 8L, 10L, 6L, 8L, 8L, 1L, 7L, 2L, 3L, 
    7L, 9L), mean_call_duration = c(57.8043647700702, 222.806, 
    132.73, 74.976645858206, 204.53, 138.8385, 138.21, 113.478, 
    162.656, 127.714, 145.507732189148, 168.676, 73.928), median_call_duration = c(29, 
    78, 25.6666666666667, 29, 36, 23.875, 23.5, 25, 44, 14, 30, 
    46, 16), max_call_duration = c(2117, 4589, 5137, 4470, 3966, 
    5137, 5137, 3249, 5137, 7201, 7201, 5137, 1941), min_call_duration = c(0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)), .Names = c("dt", "uuid_lev", 
"mean_call_duration", "median_call_duration", "max_call_duration", 
"min_call_duration"), class = c("grouped_df", "tbl_df", "tbl", 
"data.frame"), row.names = c(NA, -13L), vars = "dt", drop = TRUE, indices = list(
    0:3, 4:5, 6L, 7:8, 9:12), group_sizes = c(4L, 2L, 1L, 2L, 
4L), biggest_group_size = 4L, labels = structure(list(dt = structure(c(17702, 
17703, 17704, 17705, 17706), class = "Date")), class = "data.frame", row.names = c(NA, 
-5L), vars = "dt", drop = TRUE, .Names = "dt"))

这是我的比例函数：

scale_0_1 <- function(x) {

  return((x - min(x)) /(max(x) - min(x)))

}

当我在以下各列上应用该函数时，它将起作用：

mean_call_duration
<dbl>
median_call_duration
<dbl>
max_call_duration

但是当我使用以下方法来应用它时：

call_logs_call_duration_stats_agg %>% 
  mutate(mean_call_duration = scale_0_1(mean_call_duration),
         median_call_duration = scale_0_1(median_call_duration),
         max_call_duration = scale_0_1(max_call_duration))

我得到NaN：

structure(list(dt = structure(c(17702, 17702, 17702, 17702, 17703, 
17703, 17704, 17705, 17705, 17706, 17706, 17706, 17706), class = "Date"), 
    uuid_lev = c(4L, 5L, 8L, 10L, 6L, 8L, 8L, 1L, 7L, 2L, 3L, 
    7L, 9L), mean_call_duration = c(0, 1, 0.454090258714836, 
    0.104073399419383, 1, 0, NaN, 0, 1, 0.567674251699244, 0.755474861623972, 
    1, 0), median_call_duration = c(0.0636942675159236, 1, 0, 
    0.0636942675159236, 1, 0, NaN, 0, 1, 0, 0.5, 1, 0.0625), 
    max_call_duration = c(0, 0.818543046357616, 1, 0.779139072847682, 
    0, 1, NaN, 0, 1, 1, 1, 0.607604562737643, 0), min_call_duration = c(0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)), .Names = c("dt", "uuid_lev", 
"mean_call_duration", "median_call_duration", "max_call_duration", 
"min_call_duration"), class = c("grouped_df", "tbl_df", "tbl", 
"data.frame"), row.names = c(NA, -13L), vars = "dt", labels = structure(list(
    dt = structure(c(17702, 17703, 17704, 17705, 17706), class = "Date")), class = "data.frame", row.names = c(NA, 
-5L), vars = "dt", drop = TRUE, .Names = "dt"), indices = list(
    0:3, 4:5, 6L, 7:8, 9:12), drop = TRUE, group_sizes = c(4L, 
2L, 1L, 2L, 4L), biggest_group_size = 4L)

请告知mutate有什么问题吗？

Answer 1

变异没有错。由于已将data.frame分组，因此您每天都在进行缩放。第7行是1个群组，因为它只有一个日期，即2018-06-22。这意味着max和min相同，并且您被0除。因此，此行上的NaN。

如果您不想每天进行扩展，则需要在ungroup之前致电mutate，如下所示。

call_logs_call_duration_stats_agg %>% 
  ungroup() %>% 
  mutate(mean_call_duration = scale_0_1(mean_call_duration),
         median_call_duration = scale_0_1(median_call_duration),
         max_call_duration = scale_0_1(max_call_duration))


# A tibble: 13 x 6
   dt         uuid_lev mean_call_duration median_call_duration max_call_duration min_call_duration
   <date>        <int>              <dbl>                <dbl>             <dbl>             <dbl>
 1 2018-06-20        4             0                    0.234             0.0335                 0
 2 2018-06-20        5             1                    1                 0.503                  0
 3 2018-06-20        8             0.454                0.182             0.608                  0
 4 2018-06-20       10             0.104                0.234             0.481                  0
 5 2018-06-21        6             0.889                0.344             0.385                  0
 6 2018-06-21        8             0.491                0.154             0.608                  0
 7 2018-06-22        8             0.487                0.148             0.608                  0
 8 2018-06-23        1             0.337                0.172             0.249                  0
 9 2018-06-23        7             0.635                0.469             0.608                  0
10 2018-06-24        2             0.424                0                 1                      0
11 2018-06-24        3             0.532                0.25              1                      0
12 2018-06-24        7             0.672                0.5               0.608                  0
13 2018-06-24        9             0.0977               0.0312            0                      0

变异动词会产生NaN，而R则不会

1 个答案: