通过超过50个点的所有数据点传递平滑线

Question

我的数据看起来像：

year  mean.streak
1958  2.142857
1959  3.066667
1960  2.166667
1961  2.190476

我的带有本地化回归的情节的代码如下：

ggplot(aes(x = year, y = mean.streak, color = year), data = streaks)+
  geom_point(color = 'black')+
  geom_smooth(method = 'loess')

和输出：

我想通过在所有数据点中传递平滑线而不是典型的锯齿状geom_line来捕获数据的某种正弦模式。我写了多边形插值：

ggplot(df)+
  geom_point(aes(x = year, y = mean.streak, colour = year), size = 3) +
  stat_smooth(aes(x = year, y = mean.streak), method = "lm",
              formula = y ~ poly(x, 57), se = FALSE)

取自this thread。但是我得到了错误：

Warning message:
Computation failed in `stat_smooth()`:
'degree' must be less than number of unique points 

似乎因为数据点太多，this answer似乎表明了这一点。

有没有办法用59个数据点通过所有数据的平滑线？

完整数据是：

structure(list(year = 1958:2016, mean.streak = c(2.14285714285714, 
3.06666666666667, 2.16666666666667, 2.19047619047619, 2.35, 2.42857142857143, 
2.28571428571429, 1.92592592592593, 1.69230769230769, 2.61111111111111, 
3, 2.94117647058824, 2.2, 2.5, 2.13636363636364, 1.76923076923077, 
1.36111111111111, 1.41176470588235, 1.76, 2, 2.63157894736842, 
2.08695652173913, 2.86666666666667, 2.125, 3, 3.125, 2.57894736842105, 
1.84, 1.46666666666667, 1.7037037037037, 1.625, 1.67741935483871, 
1.84, 1.6, 3, 3.11111111111111, 3.66666666666667, 4.18181818181818, 
2.85714285714286, 3.66666666666667, 2.66666666666667, 2.92857142857143, 
3.1875, 2.76923076923077, 5.375, 5.18181818181818, 4.08333333333333, 
6.85714285714286, 2.77777777777778, 2.76470588235294, 3.15384615384615, 
3.83333333333333, 3.06666666666667, 3.07692307692308, 4.41666666666667, 
4.9, 5.22222222222222, 5, 5.27272727272727), median.streak = c(1, 
3, 1.5, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 1, 1, 1, 1, 2, 
2, 3, 2, 2, 2.5, 2, 2, 1, 1, 1, 1, 1, 1, 1.5, 2, 4, 4, 1, 3, 
2, 2.5, 2, 2, 5.5, 4, 2.5, 9, 2, 2, 2, 1.5, 2, 3, 2.5, 4.5, 4, 
5, 4), max.streak = c(6, 6, 9, 7, 5, 5, 7, 4, 3, 7, 9, 7, 6, 
6, 6, 4, 3, 4, 4, 10, 8, 6, 6, 5, 10, 8, 5, 6, 3, 4, 4, 4, 4, 
5, 8, 8, 11, 8, 8, 11, 10, 5, 12, 7, 10, 12, 12, 10, 7, 10, 10, 
14, 9, 7, 9, 12, 10, 14, 12), mean.std = c(-0.73762950487994, 
-0.480997734887942, -0.517355702126398, -0.387678832192802, -0.315808940316265, 
-0.455313725347534, -0.520453518496716, -0.598412265824216, -0.523171795723798, 
-0.62285788065637, -0.54170040191883, -0.590289727314622, -0.468222025966258, 
-0.639180735884434, -0.656427002478427, -0.565745564840106, -0.473399411312895, 
-0.564475310127763, -0.493531273810312, -0.543209721496256, -0.640240670332106, 
-0.510337503791441, -0.596096374402028, -0.504696265560619, -0.620412635042488, 
-0.497008319856979, -0.546623513153538, -0.613345407826292, -0.564945850817486, 
-0.581770706442245, -0.5709080560492, -0.627986564445679, -0.680973485641403, 
-0.548092447365696, -0.554620596559388, -0.483847268000936, -0.67619820292833, 
-0.613245144944101, -0.509832316970819, -0.302654541906113, -0.623276311320811, 
-0.431421947082012, -0.525548788393688, -0.244995094473986, -0.412444188256097, 
-0.112114155982405, -0.299486359079708, -0.300201791042539, -0.240281366191648, 
-0.359719754440627, -0.511417389357902, -0.474906675611613, -0.312106332395495, 
-0.449137693833681, -0.526248555772371, -0.56052848268042, -0.390017880007091, 
-0.537267264953157, -0.444528236868953)), class = c("tbl_df", 
"tbl", "data.frame"), .Names = c("year", "mean.streak", "median.streak", 
"max.streak", "mean.std"), row.names = c(NA, -59L))

Answer 1

调整范围：

ggplot(aes(x = year, y = mean.streak, color = year), data = streaks)+
  geom_point(color = 'black')+
  stat_smooth(method = 'loess', span = 0.3)

或使用样条线：

library(splines)
ggplot(aes(x = year, y = mean.streak, color = year), data = streaks)+
  geom_point(color = 'black')+
  stat_smooth(method = 'lm', formula = y ~ ns(x, 10))

通常，您不希望拟合极高次多项式。这种看起来很糟糕。将实际时间序列模型拟合到您的数据会好得多：

library(forecast)
library(zoo)

ggplot(aes(x = year, y = mean.streak, color = year), data = streaks)+
  geom_point(color = 'black')+
  geom_line(data = data.frame(year = sort(streaks$year),
                              mean.streak = fitted(auto.arima(zoo(streaks$mean.streak, 
                                                                  order.by = streaks$year)))), 
            show.legend = FALSE)