是否可以使用中位數來聚合或彙總R中的數據集？

-3

d <- aggregate(c(d$user_reported_percent, d$machine_percent), 
         by = list(d$day), FUN=median, simplify = TRUE, drop = TRUE)

但是R一直抱怨，我不確定它是否有意義與中位數聚合。

一些錯誤的R給我：在aggregate.data.frame 錯誤（as.data.frame（X），...）：參數必須具有相同的長度

然後我試圖使用發生變異，至少找位

d <- d %>% group_by(day) %>% mutate(median=median(user_reported_percent))

錯誤是：錯誤：無效標型「整數」

我希望得到任何幫助！非常感謝！

P.S均值爲一切工作完全正常

我的數據集是這樣的：

structure(list(esmFollValue = c(36.00852, 8.688648, 0.6372048, 
13.7394, 0.7599012, 16.43628, 7.569684, 0.4502016, 0.7630464, 
0.781386, 0.5116056, 0.858756, 18.06108, 0.5473332, 14.62944, 
14.62944, 14.07216, 0.5366868, 14.12892, 0.7354944), esmHappValue = c(100L, 
80L, 80L, 80L, 60L, 80L, 60L, 60L, 80L, 60L, 100L, 60L, 80L, 
60L, 60L, 60L, 60L, 60L, 80L, 60L), deviceId = structure(c(11L, 
11L, 11L, 6L, 3L, 15L, 3L, 3L, 15L, 3L, 15L, 15L, 15L, 15L, 3L, 
3L, 15L, 3L, 9L, 9L), .Label = c("1e6c1183-af64-4860-b2d6-533cab7afe6c", 
"34209e3d-1a82-4f75-95c8-846be8a1be03", "7066f4af-82f3-4369-8f45-70d1ea3d22f2", 
"7cf78328-60c5-4564-9dd0-309cb0b3d5ad", "95b11f22-91e8-46d0-88d9-4f197267aa29", 
"a0c89d2a-d22d-41d0-a070-b9887d911953", "cde8cc10-7212-4a41-ae9b-bbeb51dbe8ed", 
"d150bfa4-0b52-47a0-b450-1eb21aaada53", "d41db7bc-2b81-4111-9b32-a0aab55cb25a", 
"d7e8e8c7-5190-4f0b-aa49-72e520bc9aad", "dd1218a2-4e67-4cbf-bf4d-9e288865aa63", 
"f093abf9-22e1-47e6-ae5d-1238629d8542", "fae0dd29-2b89-4c1d-b5ad-7858abe122ac", 
"feeb0ab0-7d13-4a5c-b0df-58dd85c7f607", "ff883e61-c9a9-4e6b-8b6b-cab3e5535879" 
), class = "factor"), timestamp = c(1457272936.882, 1457337998.931, 
1457424251.996, 1457429767.632, 1457597635.755, 1457683537.604, 
1457861178.161, 1457964712.356, 1458029223.54, 1458046931.652, 
1458051135.219, 1458115293.069, 1458133652.503, 1458202019.302, 
1458203945.674, 1458203945.787, 1458306790.803, 1458308783.441, 
1458460903.755, 1458480932.088), group = structure(c(1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L), .Label = c("groupA", "groupB", "groupC", "groupD"), class = "factor"), 
    cameraFeed = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("Non-visible camera feed", 
    "Visible camera feed"), class = "factor"), timegroup = structure(c(2L, 
    1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 2L, 2L, 1L, 2L, 1L, 1L, 1L, 
    2L, 2L, 1L, 2L), .Label = c("Day", "Evening"), class = "factor"), 
    day = structure(c(4L, 2L, 6L, 6L, 5L, 1L, 4L, 2L, 6L, 6L, 
    6L, 7L, 7L, 5L, 5L, 5L, 1L, 1L, 4L, 4L), .Label = c("Friday", 
    "Monday", "Saturday", "Sunday", "Thursday", "Tuesday", "Wednesday" 
    ), class = "factor"), user_reported_percent = c(83.3333333333333, 
    66.6666666666667, 66.6666666666667, 66.6666666666667, 50, 
    66.6666666666667, 50, 50, 66.6666666666667, 50, 83.3333333333333, 
    50, 66.6666666666667, 50, 50, 50, 50, 50, 66.6666666666667, 
    50), machine_percent = c(30.0071, 7.24054, 0.531004, 11.4495, 
    0.633251, 13.6969, 6.30807, 0.375168, 0.635872, 0.651155, 
    0.426338, 0.71563, 15.0509, 0.456111, 12.1912, 12.1912, 11.7268, 
    0.447239, 11.7741, 0.612912)), .Names = c("esmFollValue", 
"esmHappValue", "deviceId", "timestamp", "group", "cameraFeed", 
"timegroup", "day", "user_reported_percent", "machine_percent" 
), row.names = c(NA, 20L), class = "data.frame")

，我想有每天每％的一種價值。

來源

2016-11-21 Zhanna Sarsenbayeva

請顯示可重現的示例和預期輸出。 – akrun

對不起，沒有具體和描述。我現在添加了數據集快照 –

「聚合」行中至少有一個錯誤。更可能的是，第一個參數應該是'd [，c（「user_reported_percent」，「machine_percent」）]'。它不能保證它可以工作，但是你收到的錯誤來自於你的第一個參數的長度比分組變量的長度要長（你明白爲什麼？）。 – nicola

與@nicola的幫助下，我用這個：

aggregate(d[,c("user_reported_percent","machine_percent")],b‌y = list(d$day), FUN=median)

，一切工作正常。非常感謝！

來源

2016-11-21 10:57:53

是否可以使用中位數來聚合或彙總R中的數據集？

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