Question

上下文是，我有许多id和许多band的时间序列，并且包括了九个id和两个band的样本s。在这里，我们可以轻松地绘制所有id的时间序列：

library(tidyverse)
df <- structure(list(id = c(1001L, 1001L, 1001L, 1001L, 1001L, 1001L, 1001L, 1001L, 1001L, 1001L, 1002L, 1002L, 1002L, 1002L, 1002L, 1002L, 1002L, 1002L, 1002L, 1002L, 1004L, 1004L, 1004L, 1004L, 1004L, 1004L, 1004L, 1004L, 1004L, 1004L, 1005L, 1005L, 1005L, 1005L, 1005L, 1005L, 1005L, 1005L, 1005L, 1005L, 1007L, 1007L, 1007L, 1007L, 1007L, 1007L, 1007L, 1007L, 1007L, 1007L, 1009L, 1009L, 1009L, 1009L, 1009L, 1009L, 1009L, 1009L, 1009L, 1009L, 1010L, 1010L, 1010L, 1010L, 1010L, 1010L, 1010L, 1010L, 1010L, 1010L, 1011L, 1011L, 1011L, 1011L, 1011L, 1011L, 1011L, 1011L, 1011L, 1011L, 1013L, 1013L, 1013L, 1013L, 1013L, 1013L, 1013L, 1013L, 1013L, 1013L), date = structure(c(1488884400, 1490612474, 1507460497, 1502276146, 1514372627, 1512644789, 1500980863, 1503572707, 1513940711, 1496660730, 1495796861, 1512644789, 1488884400, 1504436115, 1502276146, 1495796118, 1494068453, 1504868786, 1513940711, 1511780307, 1511348810, 1503572707, 1497524848, 1507028336, 1491476744, 1503572707, 1492340161, 1501844755, 1505300762, 1503140790, 1509620381, 1488884400, 1487156167, 1510052273, 1491476744, 1494068453, 1513940711, 1489748810, 1498388749, 1509620381, 1500980120, 1511780307, 1502708860, 1489748810, 1501412778, 1504436115, 1495796861, 1493204748, 1510484382, 1487156167, 1508324436, 1500548201, 1513940711, 1505732183, 1490612474, 1496660730, 1511348810, 1514372627, 1494068453, 1510052273, 1500548201, 1513076347, 1508756553, 1510484382, 1504436858, 1504004193, 1494932749, 1508324436, 1512644789, 1504868786, 1507460497, 1504004193, 1503140790, 1500980120, 1512212632, 1491476744, 1513940711, 1508756553, 1504436115, 1490612474, 1495796861, 1509188631, 1508756553, 1486292805, 1504004193, 1498388749, 1495796861, 1486292805, 1513940711, 1499684790), class = c("POSIXct", "POSIXt"), tzone = "UTC"), band = c("fit1", "fit1", "fit1", "fit1", "fit1", "fit5", "fit5", "fit5", "fit5", "fit5", "fit1", "fit1", "fit1", "fit1", "fit1", "fit5", "fit5", "fit5", "fit5", "fit5", "fit1", "fit1", "fit1", "fit1", "fit1", "fit5", "fit5", "fit5", "fit5", "fit5", "fit1", "fit1", "fit1", "fit1", "fit1", "fit5", "fit5", "fit5", "fit5", "fit5", "fit1", "fit1", "fit1", "fit1", "fit1", "fit5", "fit5", "fit5", "fit5", "fit5", "fit1", "fit1", "fit1", "fit1", "fit1", "fit5", "fit5", "fit5", "fit5", "fit5", "fit1", "fit1", "fit1", "fit1", "fit1", "fit5", "fit5", "fit5", "fit5", "fit5", "fit1", "fit1", "fit1", "fit1", "fit1", "fit5", "fit5", "fit5", "fit5", "fit5", "fit1", "fit1", "fit1", "fit1", "fit1", "fit5", "fit5", "fit5", "fit5", "fit5"), value = c(0.496538754230172, 0.503271496428091, 0.97387311299285, 0.580658673638122, 0.55924511798107, 0.832069876834949, 0.669456383223215, 1.12835570514478, 0.650077806710299, 0.380956367547047, 0.315803532869213, 0.792491389890908, 0.542150595815071, 1.03016500582205, 0.761751198659722, 0.367933240661702, 0.478285303617102, 1.68901870452092, 0.740965064159661, 1.09028738312622, 0.822334909416119, 0.758342181009204, 0.404208383270466, 0.892795714415756, 0.452540219822814, 1.15220190981348, 0.522093412373678, 0.953592910857701, 1.27850667816495, 1.10756222303339, 0.722797148902218, 0.465842402588039, 0.524130056243481, 0.724757971315511, 0.401849347220063, 0.455169211763473, 0.736683498842155, 0.530595901306756, 0.598435246507131, 0.855911625573028, 0.459872179640563, 0.851473466057886, 0.600348304937791, 0.484896112230185, 0.491357621589034, 1.21884821937325, 0.408355867626313, 0.541537217668289, 1.20173675518489, 0.61126928681528, 1.02122136799224, 0.489289990779144, 0.829092258901136, 0.88152853467569, 0.528559966420024, 0.544164467022259, 1.15093592993106, 0.876559089290843, 0.582149928218707, 1.26592404446571, 0.479960992971744, 0.840894959543198, 1.00459298341354, 0.98285777345435, 0.754965044767638, 1.14971147250154, 0.678568628236206, 1.38981008816777, 0.989354634818581, 1.25116433808614, 1.2142398253614, 1.03201975237089, 0.928602154928637, 0.642961745200205, 0.842888403466734, 0.649606669375906, 0.724490820076092, 1.68294181717141, 1.83216850101507, 0.69741924948021, 0.268972923828825, 1.16584414990533, 1.20604228862346, 0.586060027904748, 1.16356144256577, 0.52670838257608, 0.382147314320451, 0.668308513834733, 0.78509264848017, 0.733357618207109)), row.names = c(NA, -90L), class = c("grouped_df", "tbl_df", "tbl", "data.frame"), vars = c("id", "band"), drop = TRUE, indices = list(0:4, 5:9, 10:14, 15:19, 20:24, 25:29, 30:34, 35:39, 40:44, 45:49, 50:54, 55:59, 60:64, 65:69, 70:74, 75:79, 80:84, 85:89), group_sizes = c(5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L), biggest_group_size = 5L, labels = structure(list(id = c(1001L, 1001L, 1002L, 1002L, 1004L, 1004L, 1005L, 1005L, 1007L, 1007L, 1009L, 1009L, 1010L, 1010L, 1011L, 1011L, 1013L, 1013L), band = c("fit1", "fit5", "fit1", "fit5", "fit1", "fit5", "fit1", "fit5", "fit1", "fit5", "fit1", "fit5", "fit1", "fit5", "fit1", "fit5", "fit1", "fit5")), row.names = c(NA, -18L), class = "data.frame", vars = c("id", "band"), drop = TRUE, indices = list(0:4, 5:9, 10:14, 15:19, 20:24, 25:29, 30:34, 35:39, 40:44, 45:49, 50:54, 55:59, 60:64, 65:69, 70:74, 75:79, 80:84, 85:89), group_sizes = c(5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L), biggest_group_size = 5L, labels = structure(list(merge_id = c(1001L, 1001L, 1002L, 1002L, 1004L, 1004L, 1005L, 1005L, 1007L, 1007L, 1009L, 1009L, 1010L, 1010L, 1011L, 1011L, 1013L, 1013L), band = c("fit1", "fit5", "fit1", "fit5", "fit1", "fit5", "fit1", "fit5", "fit1", "fit5", "fit1", "fit5", "fit1", "fit5", "fit1", "fit5", "fit1", "fit5")), row.names = c(NA, -18L), class = "data.frame", vars = c("merge_id", "band"), drop = TRUE)))

ggplot(df, aes(x = date, y = value, colour = band)) +
  geom_point() + 
  geom_line() +
  facet_wrap(~id)

但是，当id过多时，这变得笨拙，并且图也变得太小，因此我想目视检查随机子集。我希望以下内容仅返回id中的三个，但是相反，我们得到四个id，而我们甚至没有得到每个{{1} }。我在这里选择种子1234，但是如果继续使用不同的种子，不同的band-id组合重新运行，则会得到不同的结果。

band

请注意，如果我在id调用之外进行采样，则可以使用。（这将是理想的结果）

set.seed(1234)
ggplot(
  data = df %>% filter(id %in% sample(unique(df$id), 3)), # filtering to subset of 3 ids
  mapping = aes(x = date, y = value, colour = band)
) +
  geom_point() +
  geom_line() +
  facet_wrap(~id)

为什么会这样？我看不到两个选项在逻辑上的区别。它肯定与ggplot()有关，而不与set.seed(1234) some_ids <- sample(unique(df$id), 3) # moved sample() outside of ggplot() ggplot( data = df %>% filter(id %in% some_ids), mapping = aes(x = date, y = value, colour = band) ) + geom_point() + geom_line() + facet_wrap(~id)部分有关，因为您可以将其替换为sample并仍然可以解决问题。我也意识到这可能与我的特定数据有关，因为我确实尝试使用内置数据集进行类似的reprex，但是我无法想象那会是什么，因为这已经是一个相当有限的子集了。 >

编辑：例如，如果我使用此简化后的数据集，则无法重现此错误。我很困惑，因为除了实际值之外，我无法分辨此数据集与unique(df$id)中的数据集之间的任何区别。

c(1001, 1002, 1004, 1005, 1007, 1009, 1010, 1011, 1013)

Answer 1

TLDR：过滤器表达式会被多次求值，因此您不应使用非确定性表达式。

不确定这是否足以解决问题，但是如果尝试使用不同的种子运行示例，则会注意到图表数量随每个种子而变化。这表明我们正在过滤数据帧的id的数量随每个种子而变化，这表明sample实际上被多次调用。我们可以通过创建一个代替sample的函数来确认这一点：

sample_out <- function(data, n) {
  print("running sample_out ")
  return (sample(data, n))
}

，然后用它代替sample：

ggplot(
  data = df %>% filter(id %in% sample_out(unique(df$id), 3)), 
  mapping = aes(x = date, y = value, colour = band)
)

您会看到sample_out实际上被多次调用。在我的会话中，无论种子如何，上面的数据都会被调用18次。实验不同的数据帧大小，似乎sample会被调用（row_count / 5）次。这意味着filter以某种方式多次评估其参数。一个完整的答案将解释为什么filter会发生这种情况，但是这让我有点迷惑。我相信相关资料在这里：

https://github.com/tidyverse/dplyr/blob/master/R/tbl-df.r#L55

filter.tbl_df <- function(.data, ..., .preserve = TRUE) {
  // elided
  out <- filter_impl(.data, quo)

filter_impl基本上调用了C ++实现，我认为关键是：

https://github.com/tidyverse/dplyr/blob/master/src/filter.cpp#L408

template <typename SlicedTibble>
SEXP filter_template(const SlicedTibble& gdf, const NamedQuosure& quo) {
  // elided
  Proxy call_proxy(quo.expr(), gdf, quo.env()) ;
  // elided
  int ngroups = gdf.ngroups() ;    
  // elided    
  for (int i = 0; i < ngroups; i++, ++git) {
    // elided
    LogicalVector g_test = check_result_lgl_type(call_proxy.get(indices));
    // elided
  }
  // elided
}

请注意，对于每个小节，将执行call_proxy.get。我假设我们看到sample_out被调用了18次，因为相应的小标题中有18个组。

无论如何，可以通过发布到相关的dplyr社区联系人来快速，权威地回答此问题。在关于dyplr的冒险学习中，我无法找到关于此的警告，因此可能是我遗漏了一些东西。 dplyr的文档讨论了其评估结果与https://dplyr.tidyverse.org/articles/programming.html可能有所不同。

大多数dplyr函数使用非标准评估（NSE）。这是一个笼统的术语，表示他们没有遵循通常的R评估规则。而是，它们捕获您键入的表达式并以自定义方式对其求值。

为什么过滤ggplot（）中的数据集样本会返回不正确的样本？

1 个答案: