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Question

我是R的初学者，我想向您寻求帮助。

任务：我想制作一张图表，代表一天中每小时的用水量。该图由不同日期的几条曲线组成（例如，请参见链接here）。

我将每天的数据划分为子列表：

    > head(aaa)
    [[1]]
                   by60min  consumption
    1  2018-07-01 00:05:00            0
    2  2018-07-01 01:05:00            0
    3  2018-07-01 02:05:00            0
    4  2018-07-01 03:05:00            0
    ....
    [[2]]
                   by60min  consumption
    25 2018-07-02 00:05:00            0
    26 2018-07-02 01:05:00            0
    27 2018-07-02 02:05:00            0
    28 2018-07-02 03:05:00            0

有时候，没有水消耗，我想避免将这些天绘制到图表中。在这里，我被困住了。我不知道怎么做。我的想法是删除消耗量为零的所有天数，然后绘制非零天数，但是我做不到。有什么想法怎么做（绘制非零天或/和如何从列表中删除子列表）？

非常感谢您。

Luboš

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# 1st step - tibble:
    aaa <- as.tibble(aaa)
    aaa
# A tibble: 1,487 x 2
    by60min             consumption
    <fct>                     <dbl>
    1 2018-07-01 00:05:00         0
    2 2018-07-01 01:05:00         0
    3 2018-07-01 02:05:00         0
    4 2018-07-01 03:05:00         0
    5 2018-07-01 04:05:00         0
    6 2018-07-01 05:05:00         0
    7 2018-07-01 06:05:00         0
    8 2018-07-01 07:05:00     0.101
    9 2018-07-01 08:05:00     0.167
   10 2018-07-01 09:05:00     0.267
   # ... with 1,477 more rows

# 2nd step - plot:
    aaa %>%
      mutate(day = factor(day(ymd_hms(by60min))),
             hour = factor(hour(ymd_hms(by60min)))) %>%
      group_by(day) %>%
      filter(sum(consumption) > 0) %>%
      ggplot(mapping = aes(x = hour, y = consumption, 
                           col = day, 
                           show.legend = FALSE)) +
      geom_line(show.legend = FALSE)

# OUTPUT (the picture below) - bar graph instead of line chart - why?
# please NOTE that akt_spotreba == consumption

dput(aaa) # I inserted only first three rows
structure(list(by60min = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 
7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 
20L, 21L, 22L, 23L, 24L, 25L, 26L, 27L, 28L, 29L, 30L, 31L, 32L,

Answer 1

这是一个tidyverse方法，根据您提供的内容，使用一个简单的示例数据集。

l1 = data.frame(by60min = c("2018-07-01 00:05:00","2018-07-01 01:05:00","2018-07-01 02:05:00"),
                consumption = 0)

l2 = data.frame(by60min = c("2018-07-02 00:05:00","2018-07-02 01:05:00","2018-07-02 02:05:00"),
                consumption = c(0,2,30))

l3 = data.frame(by60min = c("2018-07-03 00:05:00","2018-07-03 01:05:00","2018-07-03 02:05:00"),
                consumption = c(10,8,2))

l = list(l1,l2,l3)

您的原始数据如下：

[[1]]
by60min consumption
1 2018-07-01 00:05:00           0
2 2018-07-01 01:05:00           0
3 2018-07-01 02:05:00           0

[[2]]
by60min consumption
1 2018-07-02 00:05:00           0
2 2018-07-02 01:05:00           2
3 2018-07-02 02:05:00          30

[[3]]
by60min consumption
1 2018-07-03 00:05:00          10
2 2018-07-03 01:05:00           8
3 2018-07-03 02:05:00           2

library(tidyverse)
library(lubridate)

map_df(l, data.frame) %>%                         # combine list element to one dataframe
  mutate(day = factor(date(ymd_hms(by60min))),    # get day from date
         hr = hour(ymd_hms(by60min))) %>%         # get hour from date
  group_by(day) %>%                               # for each day
  filter(sum(consumption) > 0) %>%                # calculate sum of consumption and remove days where this is 0
  ungroup() %>%
  ggplot(aes(hr, consumption, col=day))+          # plot lines
  geom_line()

输出图：

在R中绘制小时曲线

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1 个答案: