我有一個 data.table test_dt，我想在其中使用lowess函式平滑 y 列。
test_dt <- structure(list(x = c(28.75, 30, 31.25, 32.5, 33.75, 35, 36.25,
37.5, 38.75, 40, 41.25, 42.5, 43.75, 45, 46.25, 47.5, 48.75,
50, 52.5, 55, 57.5, 60, 62.5, 63.75, 65, 67.5, 70, 72.5, 75,
77.5, 80, 82.5, 85, 87.5, 90, 92.5, 95, 97.5, 100, 102.5, 103.75,
105, 106.25, 107.5, 108.75, 110, 111.25, 112.5, 113.75, 115,
116.25, 117.5, 118.75, 120, 121.25, 122.5, 125, 130, 135, 140,
145), y = c(116.78, 115.53, 114.28, 113.05, 111.78, 110.53, 109.28,
108.05, 106.78, 105.53, 104.28, 103.025, 101.775, 100.525, 99.28,
98.05, 96.8, 95.525, 93.1, 90.65, 88.225, 85.775, 83.35, 82.15,
80.9, 78.5, 76.075, 73.675, 71.25, 68.85, 66.5, 64.075, 61.725,
59.4, 57.075, 54.725, 52.475, 50.225, 48, 45.75, 44.65, 43.55,
42.475, 41.45, 40.35, 39.275, 38.25, 37.225, 36.175, 35.175,
34.175, 33.225, 32.275, 31.3, 30.35, 29.45, 27.625, 24.175, 21,
18.125, 15.55), z = c(116.778248424972, 115.531456655985, 114.284502467544,
113.034850770519, 111.784500981402, 110.533319511795, 109.284500954429,
108.034850457264, 106.784502297216, 105.531265565238, 104.278221015846,
103.026780249377, 101.775992395759, 100.528761292272, 99.2853168637851,
98.043586202838, 96.8021989104315, 95.5702032427799, 93.1041279347743,
90.6575956222915, 88.2179393348852, 85.783500434839, 83.3503011023971,
82.136280706039, 80.922846825298, 78.4965179152157, 76.0823895453039,
73.6686672097464, 71.264486719796, 68.8702598156142, 66.4865368523571,
64.1182523898466, 61.7552221811808, 59.4004347738795, 57.0823289450761,
54.7908645949795, 52.5071096685879, 50.2308279167219, 47.9940967492558,
45.7658417529877, 44.6514226583931, 43.5622751034012, 42.4876666190815,
41.4173110074806, 40.3555584369672, 39.3004471381618, 38.2552969838653,
37.2202353638959, 36.1963659189447, 35.1889616530209, 34.2004259883859,
33.2295174626826, 32.2669278456991, 31.3171387914754, 30.3742375589802,
29.4555719783757, 27.6243725086786, 23.9784367995753, 27.625,
27.625, 27.625)), row.names = c(NA, -61L), class = c("data.table",
"data.frame"))
如下圖所示，我得到了意想不到的結果。預期結果是下圖中的線（z列）應緊跟點（y列）。
這是我的代碼 -
library(data.table)
library(ggplot2)
test_dt[, z := lowess(x = x, y = y, f = 0.1)$y]
ggplot(test_dt) geom_point(aes(x, y)) geom_line(aes(x, z))
一季度。有人可以建議為什么lowess不能正確平滑嗎？
Q2。由于lowess沒有按預期作業，R 中有沒有其他函式可以更有效地平滑y列而不產生尖峰（就像lowess在邊界點上那樣）？
uj5u.com熱心網友回復：
你可以loess改用：
test_dt[, z := predict(loess(y ~ x, data = test_dt))]
ggplot(test_dt) geom_point(aes(x, y)) geom_line(aes(x, z))
但請注意，如果您只想繪制線條，這正是使用的方法geom_smooth，因此甚至無需創建z列，您可以執行以下操作：
ggplot(test_dt, aes(x, y)) geom_point() geom_smooth()
#> `geom_smooth()` using method = 'loess' and formula 'y ~ x'
由reprex 包( v2.0.0 )于 2021 年 11 月 7 日創建
uj5u.com熱心網友回復：
通過將迭代次數保持為零來解決問題。lowess就像loess迭代保持為零一樣。
test_dt[, z := lowess(x = x, y = y, f = 0.1, iter=0)$y]
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