pivot_widerの問題「 `values_from`の値は一意に識別されません。出力にはリスト列が含まれます」

Question

私のデータは次のようになります：

# 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

pivot_wider関数を使用して、name列に基づいてデータを展開しようとしています。次のコードを使用します。

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

次の警告メッセージが表示されます。

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

出力は次のようになります。

 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>

ここが問題に言及し、それを解決するために要約統計量を使用することを提案していることを知っています。しかし、私は時系列データを持っているため、毎日は単一の値（複数の値ではない）を持っているので、要約統計を使用したくありません。

問題はval列に重複があるためです（つまり、上記の例では7が3回発生しています）。

この問題をpivot_widerして克服する方法に関する提案はありますか？

データ：

 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))

Ronak Shah · Accepted Answer

nameごとに一意の識別子の行を作成し、pivot_widerを使用します

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

Ameer · Answer

通常、エラー

Warning message: Values in `val` are not uniquely identified; output will contain list-cols.

ほとんどの場合、（val列を除外した後の）データの重複行によって引き起こされ、val列の重複ではありません。

which(duplicated(d)) # [1] 14 65

OPのデータには2つの重複する行があるため、この問題が発生しています。重複する行を削除すると、エラーも解消されます。

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

# A tibble: 50 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 40 more rows

Daniel Antal · Answer

この問題は、より広く拡散/ピボットしたいデータの識別子が重複しているために発生します。上記の両方の提案、つまりmutate(row = row_number())を使用して行番号から一意の人工IDを作成するか、distinct行のみをフィルタリングすると、ピボットを広げることができますが、テーブルの構造が変更されます。論理的で組織的な問題が発生する可能性が高く、次に何かに参加しようとすると問題が発生します。

id_colsパラメータを明示的に使用して、ピボットワイド後にピボットを一意にする必要があることを確認し、問題が発生した場合は、最初に元のテーブルを再編成することをお勧めします。もちろん、異なる行にフィルタリングしたり、新しいIDを追加したりする理由が見つかるかもしれませんが、おそらくコードの早い段階での重複を避けたいでしょう。