传播与传播
Spread vs dcast
我有一个这样的table，
> head(dt2)
Weight Height Fitted interval limit value
1 65.6 174.0 71.91200 pred lwr 53.73165
2 80.7 193.5 91.63237 pred lwr 73.33198
3 72.6 186.5 84.55326 pred lwr 66.31751
4 78.8 187.2 85.26117 pred lwr 67.02004
5 74.8 181.5 79.49675 pred lwr 61.29244
6 86.4 184.0 82.02501 pred lwr 63.80652
我希望它是这样的，
> head(reshape2::dcast(dt2,
Weight + Height + Fitted + interval ~ limit,
fun.aggregate = mean))
Weight Height Fitted interval lwr upr
1 42.0 153.4 51.07920 conf 49.15463 53.00376
2 42.0 153.4 51.07920 pred 32.82122 69.33717
3 43.2 160.0 57.75378 conf 56.35240 59.15516
4 43.2 160.0 57.75378 pred 39.54352 75.96404
5 44.8 149.5 47.13512 conf 44.87642 49.39382
6 44.8 149.5 47.13512 pred 28.83891 65.43133
但是使用 tidyr::spread，我该怎么做？
我在用，
> tidyr::spread(dt2, limit, value)
但是出现错误，
Error: Duplicate identifiers for rows (1052, 1056), (238, 242), (1209, 1218), (395, 404), (839, 1170), (25, 356), (1173, 1203, 1215), (359, 389, 401), (1001, 1200), (187, 386), (906, 907), (92, 93), (930, 1144), (116, 330), (958, 1171), (144, 357), (902, 1018), (88, 204), (960, 1008), (146, 194), (1459, 1463), (645, 649), (1616, 1625), (802, 811), (1246, 1577), (432, 763), (1580, 1610, 1622), (766, 796, 808), (1408, 1607), (594, 793), (1313, 1314), (499, 500), (1337, 1551), (523, 737), (1365, 1578), (551, 764), (1309, 1425), (495, 611), (1367, 1415), (553, 601)
随机 10 行::
> dt[sample(nrow(dt), 10), ]
Weight Height Fitted interval limit value
1253 52.2 162.5 60.28203 conf upr 61.51087
426 49.1 158.8 56.54022 pred upr 74.75756
1117 78.4 184.5 82.53066 conf lwr 80.98778
1171 85.9 166.4 64.22611 conf lwr 63.21254
948 61.4 177.8 75.75494 conf lwr 74.66393
384 90.9 172.7 70.59731 pred lwr 52.41828
289 75.9 172.7 70.59731 pred lwr 52.41828
3 44.8 149.5 47.13512 pred lwr 28.83891
774 87.3 182.9 80.91258 pred upr 99.12445
772 86.4 175.3 73.22669 pred upr 91.40919
假设您开始时使用的数据如下所示：
mydf
# Weight Height Fitted interval limit value
# 1 42 153.4 51.0792 conf lwr 49.15463
# 2 42 153.4 51.0792 pred lwr 32.82122
# 3 42 153.4 51.0792 conf upr 53.00376
# 4 42 153.4 51.0792 pred upr 69.33717
# 5 42 153.4 51.0792 conf lwr 60.00000
# 6 42 153.4 51.0792 pred lwr 90.00000
请注意分组列（1 到 5）的第 5 行和第 6 行中的重复项。这基本上就是 "tidyr" 告诉你的。第一行和第五行是重复的，第二行和第六行也是。
tidyr::spread(mydf, limit, value)
# Error: Duplicate identifiers for rows (1, 5), (2, 6)
正如@Jaap 所建议的，解决方案是首先 "summarise" 数据。由于 "tidyr" 仅用于重塑数据（不像 "reshape2"，它聚合和重塑），您需要在更改数据形式之前使用 "dplyr" 执行聚合。在这里，我在 "value" 列中使用 summarise 完成了此操作。
如果你在summarise步停止执行，你会发现我们原来的6行数据集有"shrunk"到4行。现在，spread 将按预期工作。
mydf %>%
group_by(Weight, Height, Fitted, interval, limit) %>%
summarise(value = mean(value)) %>%
spread(limit, value)
# Source: local data frame [2 x 6]
#
# Weight Height Fitted interval lwr upr
# (dbl) (dbl) (dbl) (chr) (dbl) (dbl)
# 1 42 153.4 51.0792 conf 54.57731 53.00376
# 2 42 153.4 51.0792 pred 61.41061 69.33717
这与 dcast 和 fun.aggregate = mean 的预期输出相匹配。
reshape2::dcast(mydf, Weight + Height + Fitted + interval ~ limit, fun.aggregate = mean)
# Weight Height Fitted interval lwr upr
# 1 42 153.4 51.0792 conf 54.57731 53.00376
# 2 42 153.4 51.0792 pred 61.41061 69.33717
示例数据：
mydf <- structure(list(Weight = c(42, 42, 42, 42, 42, 42), Height = c(153.4,
153.4, 153.4, 153.4, 153.4, 153.4), Fitted = c(51.0792, 51.0792,
51.0792, 51.0792, 51.0792, 51.0792), interval = c("conf", "pred",
"conf", "pred", "conf", "pred"), limit = structure(c(1L, 1L,
2L, 2L, 1L, 1L), .Label = c("lwr", "upr"), class = "factor"),
value = c(49.15463, 32.82122, 53.00376, 69.33717, 60,
90)), .Names = c("Weight", "Height", "Fitted", "interval",
"limit", "value"), row.names = c(NA, 6L), class = "data.frame")
这里有 data.table 个替代 dplyr 的选项。使用 Ananda 的回答中的 mydf。
library(data.table)
library(magrittr)
library(tidyr)
DT <- data.table(mydf)
首先，您可以使用 by 计算每个限制的平均值。
DT[, .(lwr = mean(value[limit == "lwr"]),
upr = mean(value[limit == "upr"])),
by = .(Weight, Height, Fitted, interval)]
如果这个limit == ...看起来硬编码太多，可以先聚合成长格式，然后spread。这是有效的，因为一旦你聚合，就没有重复。
DT[, .(value = mean(value)), by = .(Weight, Height, Fitted, interval, limit)] %>%
spread(key = "limit", value = "value")
两者都得到你
# Weight Height Fitted interval lwr upr
#1: 42 153.4 51.0792 conf 54.57731 53.00376
#2: 42 153.4 51.0792 pred 61.41061 69.33717
我有一个这样的table，
> head(dt2)
Weight Height Fitted interval limit value
1 65.6 174.0 71.91200 pred lwr 53.73165
2 80.7 193.5 91.63237 pred lwr 73.33198
3 72.6 186.5 84.55326 pred lwr 66.31751
4 78.8 187.2 85.26117 pred lwr 67.02004
5 74.8 181.5 79.49675 pred lwr 61.29244
6 86.4 184.0 82.02501 pred lwr 63.80652
我希望它是这样的，
> head(reshape2::dcast(dt2,
Weight + Height + Fitted + interval ~ limit,
fun.aggregate = mean))
Weight Height Fitted interval lwr upr
1 42.0 153.4 51.07920 conf 49.15463 53.00376
2 42.0 153.4 51.07920 pred 32.82122 69.33717
3 43.2 160.0 57.75378 conf 56.35240 59.15516
4 43.2 160.0 57.75378 pred 39.54352 75.96404
5 44.8 149.5 47.13512 conf 44.87642 49.39382
6 44.8 149.5 47.13512 pred 28.83891 65.43133
但是使用 tidyr::spread，我该怎么做？
我在用，
> tidyr::spread(dt2, limit, value)
但是出现错误，
Error: Duplicate identifiers for rows (1052, 1056), (238, 242), (1209, 1218), (395, 404), (839, 1170), (25, 356), (1173, 1203, 1215), (359, 389, 401), (1001, 1200), (187, 386), (906, 907), (92, 93), (930, 1144), (116, 330), (958, 1171), (144, 357), (902, 1018), (88, 204), (960, 1008), (146, 194), (1459, 1463), (645, 649), (1616, 1625), (802, 811), (1246, 1577), (432, 763), (1580, 1610, 1622), (766, 796, 808), (1408, 1607), (594, 793), (1313, 1314), (499, 500), (1337, 1551), (523, 737), (1365, 1578), (551, 764), (1309, 1425), (495, 611), (1367, 1415), (553, 601)
随机 10 行::
> dt[sample(nrow(dt), 10), ]
Weight Height Fitted interval limit value
1253 52.2 162.5 60.28203 conf upr 61.51087
426 49.1 158.8 56.54022 pred upr 74.75756
1117 78.4 184.5 82.53066 conf lwr 80.98778
1171 85.9 166.4 64.22611 conf lwr 63.21254
948 61.4 177.8 75.75494 conf lwr 74.66393
384 90.9 172.7 70.59731 pred lwr 52.41828
289 75.9 172.7 70.59731 pred lwr 52.41828
3 44.8 149.5 47.13512 pred lwr 28.83891
774 87.3 182.9 80.91258 pred upr 99.12445
772 86.4 175.3 73.22669 pred upr 91.40919
假设您开始时使用的数据如下所示：
mydf
# Weight Height Fitted interval limit value
# 1 42 153.4 51.0792 conf lwr 49.15463
# 2 42 153.4 51.0792 pred lwr 32.82122
# 3 42 153.4 51.0792 conf upr 53.00376
# 4 42 153.4 51.0792 pred upr 69.33717
# 5 42 153.4 51.0792 conf lwr 60.00000
# 6 42 153.4 51.0792 pred lwr 90.00000
请注意分组列（1 到 5）的第 5 行和第 6 行中的重复项。这基本上就是 "tidyr" 告诉你的。第一行和第五行是重复的，第二行和第六行也是。
tidyr::spread(mydf, limit, value)
# Error: Duplicate identifiers for rows (1, 5), (2, 6)
正如@Jaap 所建议的，解决方案是首先 "summarise" 数据。由于 "tidyr" 仅用于重塑数据（不像 "reshape2"，它聚合和重塑），您需要在更改数据形式之前使用 "dplyr" 执行聚合。在这里，我在 "value" 列中使用 summarise 完成了此操作。
如果你在summarise步停止执行，你会发现我们原来的6行数据集有"shrunk"到4行。现在，spread 将按预期工作。
mydf %>%
group_by(Weight, Height, Fitted, interval, limit) %>%
summarise(value = mean(value)) %>%
spread(limit, value)
# Source: local data frame [2 x 6]
#
# Weight Height Fitted interval lwr upr
# (dbl) (dbl) (dbl) (chr) (dbl) (dbl)
# 1 42 153.4 51.0792 conf 54.57731 53.00376
# 2 42 153.4 51.0792 pred 61.41061 69.33717
这与 dcast 和 fun.aggregate = mean 的预期输出相匹配。
reshape2::dcast(mydf, Weight + Height + Fitted + interval ~ limit, fun.aggregate = mean)
# Weight Height Fitted interval lwr upr
# 1 42 153.4 51.0792 conf 54.57731 53.00376
# 2 42 153.4 51.0792 pred 61.41061 69.33717
示例数据：
mydf <- structure(list(Weight = c(42, 42, 42, 42, 42, 42), Height = c(153.4,
153.4, 153.4, 153.4, 153.4, 153.4), Fitted = c(51.0792, 51.0792,
51.0792, 51.0792, 51.0792, 51.0792), interval = c("conf", "pred",
"conf", "pred", "conf", "pred"), limit = structure(c(1L, 1L,
2L, 2L, 1L, 1L), .Label = c("lwr", "upr"), class = "factor"),
value = c(49.15463, 32.82122, 53.00376, 69.33717, 60,
90)), .Names = c("Weight", "Height", "Fitted", "interval",
"limit", "value"), row.names = c(NA, 6L), class = "data.frame")
这里有 data.table 个替代 dplyr 的选项。使用 Ananda 的回答中的 mydf。
library(data.table)
library(magrittr)
library(tidyr)
DT <- data.table(mydf)
首先，您可以使用 by 计算每个限制的平均值。
DT[, .(lwr = mean(value[limit == "lwr"]),
upr = mean(value[limit == "upr"])),
by = .(Weight, Height, Fitted, interval)]
如果这个limit == ...看起来硬编码太多，可以先聚合成长格式，然后spread。这是有效的，因为一旦你聚合，就没有重复。
DT[, .(value = mean(value)), by = .(Weight, Height, Fitted, interval, limit)] %>%
spread(key = "limit", value = "value")
两者都得到你
# Weight Height Fitted interval lwr upr
#1: 42 153.4 51.0792 conf 54.57731 53.00376
#2: 42 153.4 51.0792 pred 61.41061 69.33717