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r - ivot_wider问题“values_from”中的值未唯一标识;输出将包含list-cols”

转载 作者:行者123 更新时间:2023-12-03 08:37:22 24 4
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我的数据如下所示:

# A tibble: 6 x 4
name val time x1
<chr> <dbl> <date> <dbl>
1 C Farolillo 7 2016-04-20 51.5
2 C Farolillo 3 2016-04-21 56.3
3 C Farolillo 7 2016-04-22 56.3
4 C Farolillo 13 2016-04-23 57.9
5 C Farolillo 7 2016-04-24 58.7
6 C Farolillo 9 2016-04-25 59.0


我正在尝试使用 pivot_wider函数来扩展基于 name列的数据。我使用以下代码:

yy <- d %>% 
pivot_wider(., names_from = name, values_from = val)


这给了我以下警告信息:

Warning message:
Values in `val` are not uniquely identified; output will contain list-cols.
* Use `values_fn = list(val = list)` to suppress this warning.
* Use `values_fn = list(val = length)` to identify where the duplicates arise
* Use `values_fn = list(val = summary_fun)` to summarise duplicates


输出如下:

       time       x1        out1    out2 
2016-04-20 51.50000 <dbl> <dbl>
2 2016-04-21 56.34615 <dbl> <dbl>
3 2016-04-22 56.30000 <dbl> <dbl>
4 2016-04-23 57.85714 <dbl> <dbl>
5 2016-04-24 58.70968 <dbl> <dbl>
6 2016-04-25 58.96774 <dbl> <dbl>


我知道 here提到了此问题,并建议使用汇总统计信息解决此问题。但是我有时间序列数据,因此不想使用汇总统计信息,因为每天都有一个值(而不是多个值)。

我知道问题是因为 val列具有重复项(即在上面的示例中7次出现了3次)。

关于如何ivot_wider和克服此问题有什么建议吗?

数据:

    d <- structure(list(name = c("C Farolillo", "C Farolillo", "C Farolillo", 
"C Farolillo", "C Farolillo", "C Farolillo", "C Farolillo", "C Farolillo",
"C Farolillo", "C Farolillo", "C Farolillo", "C Farolillo", "C Farolillo",
"C Farolillo", "C Farolillo", "C Farolillo", "C Farolillo", "C Farolillo",
"C Farolillo", "C Farolillo", "C Farolillo", "C Farolillo", "C Farolillo",
"C Farolillo", "C Farolillo", "C Farolillo", "C Farolillo", "C Farolillo",
"C Farolillo", "C Farolillo", "C Farolillo", "C Farolillo", "C Farolillo",
"C Farolillo", "C Farolillo", "C Farolillo", "C Farolillo", "C Farolillo",
"C Farolillo", "C Farolillo", "C Farolillo", "C Farolillo", "C Farolillo",
"C Farolillo", "C Farolillo", "C Farolillo", "C Farolillo", "C Farolillo",
"C Farolillo", "C Farolillo", "C Farolillo", "Plaza Eliptica",
"Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica",
"Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica",
"Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica",
"Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica",
"Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica",
"Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica",
"Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica",
"Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica",
"Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica",
"Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica",
"Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica",
"Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica", "Plaza Eliptica",
"Plaza Eliptica", "Plaza Eliptica"), val = c(7, 3, 7, 13, 7,
9, 20, 19, 4, 5, 5, 2, 6, 6, 16, 13, 7, 6, 3, 3, 6, 10, 5, 3,
5, 3, 4, 4, 10, 11, 4, 13, 8, 2, 8, 10, 3, 10, 14, 4, 2, 4, 6,
6, 8, 8, 3, 3, 13, 10, 13, 32, 25, 31, 34, 26, 33, 35, 43, 22,
22, 21, 10, 33, 33, 48, 47, 27, 23, 11, 13, 25, 31, 20, 16, 10,
9, 23, 11, 23, 26, 16, 34, 17, 4, 24, 21, 10, 26, 32, 10, 5,
9, 19, 14, 27, 27, 10, 8, 28, 32, 25), time = structure(c(16911,
16912, 16913, 16914, 16915, 16916, 16917, 16918, 16919, 16920,
16921, 16922, 16923, 16923, 16924, 16925, 16926, 16927, 16928,
16929, 16930, 16931, 16932, 16933, 16934, 16935, 16936, 16937,
16938, 16939, 16940, 16941, 16942, 16943, 16944, 16945, 16946,
16947, 16948, 16949, 16950, 16951, 16952, 16953, 16954, 16955,
16956, 16957, 16958, 16959, 16960, 16911, 16912, 16913, 16914,
16915, 16916, 16917, 16918, 16919, 16920, 16921, 16922, 16923,
16923, 16924, 16925, 16926, 16927, 16928, 16929, 16930, 16931,
16932, 16933, 16934, 16935, 16936, 16937, 16938, 16939, 16940,
16941, 16942, 16943, 16944, 16945, 16946, 16947, 16948, 16949,
16950, 16951, 16952, 16953, 16954, 16955, 16956, 16957, 16958,
16959, 16960), class = "Date"), x1 = c(51.5, 56.3461538461538,
56.3, 57.8571428571429, 58.7096774193548, 58.9677419354839, 64.4615384615385,
61.9310344827586, 60.3214285714286, 59.4137931034483, 59.5806451612903,
57.3448275862069, 64.0333333333333, 64.0333333333333, 70.15625,
71.3636363636364, 62.8125, 56.4375, 56.4516129032258, 51.741935483871,
52.84375, 53.09375, 52.969696969697, 54, 54.3870967741936, 60.3870967741936,
64.4516129032258, 66.2903225806452, 68.2333333333333, 69.7741935483871,
70.5806451612903, 73.8275862068966, 72.8181818181818, 64.6764705882353,
64.4838709677419, 68.7741935483871, 62.1764705882353, 68.969696969697,
70.1935483870968, 59.6774193548387, 59.9677419354839, 63.125,
67.5882352941177, 71.4705882352941, 73.8529411764706, 76.1935483870968,
72.6451612903226, 76.0645161290323, 76.4193548387097, 81.7741935483871,
85.0645161290323, 51.5, 56.3461538461538, 56.3, 57.8571428571429,
58.7096774193548, 58.9677419354839, 64.4615384615385, 61.9310344827586,
60.3214285714286, 59.4137931034483, 59.5806451612903, 57.3448275862069,
64.0333333333333, 64.0333333333333, 70.15625, 71.3636363636364,
62.8125, 56.4375, 56.4516129032258, 51.741935483871, 52.84375,
53.09375, 52.969696969697, 54, 54.3870967741936, 60.3870967741936,
64.4516129032258, 66.2903225806452, 68.2333333333333, 69.7741935483871,
70.5806451612903, 73.8275862068966, 72.8181818181818, 64.6764705882353,
64.4838709677419, 68.7741935483871, 62.1764705882353, 68.969696969697,
70.1935483870968, 59.6774193548387, 59.9677419354839, 63.125,
67.5882352941177, 71.4705882352941, 73.8529411764706, 76.1935483870968,
72.6451612903226, 76.0645161290323, 76.4193548387097, 81.7741935483871,
85.0645161290323)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA,
-102L))

最佳答案

为每个name创建一个唯一的标识符行,然后使用pivot_wider

library(dplyr)

d %>%
group_by(name) %>%
mutate(row = row_number()) %>%
tidyr::pivot_wider(names_from = name, values_from = val) %>%
select(-row)

# A tibble: 51 x 4
# time x1 `C Farolillo` `Plaza Eliptica`
# <date> <dbl> <dbl> <dbl>
# 1 2016-04-20 51.5 7 32
# 2 2016-04-21 56.3 3 25
# 3 2016-04-22 56.3 7 31
# 4 2016-04-23 57.9 13 34
# 5 2016-04-24 58.7 7 26
# 6 2016-04-25 59.0 9 33
# 7 2016-04-26 64.5 20 35
# 8 2016-04-27 61.9 19 43
# 9 2016-04-28 60.3 4 22
#10 2016-04-29 59.4 5 22
# … with 41 more rows

关于r - ivot_wider问题“values_from”中的值未唯一标识;输出将包含list-cols”,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/58837773/

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