我想根据字符串变量的唯一值来总结一个数据帧。
df1 <- structure(list(lllocatie = structure(c(3L, 13L, 5L, 10L, 4L, 32L, 10L, 10L, 22L, 4L, 36L, 37L, 31L, 15L, 23L, 20L, 34L, 8L, 35L, 24L, 19L, 19L, 2L, 29L, 26L, 25L, 25L, 30L, 8L, 22L, 9L, 20L, 19L, 12L, 16L, 38L, 6L, 27L, 7L, 11L, 17L, 33L, 14L, 2L, 21L, 18L, 9L, 28L, 32L, 1L), .Label = c("Annen", "Appingedam", "Assen", "Eleveld", "Emmen", "Farmsum", "Froombosch", "Garrelsweer", "Garsthuizen", "Geelbroek", "Hellum", "Hoogezand", "Hooghalen", "Huizinge", "Langelo", "Leermens", "Meedhuizen", "Onderdendam", "Oosterwijtwerd", "Overschild", "Roodeschool", "Roswinkel", "Sappemeer", "Sint Annen", "Slochteren", "Startenhuizen", "Steendam", "Stitswerd", "t-Zandt", "Ten Post", "Tjuchem", "Toornwerd", "Tripscompagnie", "Westerbroek", "Westerwijtwerd", "Winneweer", "Woudbloem", "Zandeweer"), class = "factor"), lat = c(52.992, 52.928, 52.771, 52.952, 52.965, 53.358, 52.953, 52.956, 52.831, 52.961, 53.32, 53.21, 53.294, 53.084, 53.16, 53.285, 53.177, 53.305, 53.316, 53.315, 53.333, 53.336, 53.332, 53.363, 53.368, 53.208, 53.202, 53.294, 53.306, 52.833, 53.37, 53.279, 53.323, 53.17, 53.345, 53.39,
53.316, 53.275, 53.194, 53.226, 53.294, 53.156, 53.359, 53.335, 53.423, 53.324, 53.372, 53.365, 53.351, 53.061), lon = c(6.548, 6.552, 6.914, 6.575, 6.573, 6.657, 6.572, 6.562, 7.032, 6.57, 6.74, 6.747, 6.868, 6.465, 6.805, 6.795, 6.685, 6.793, 6.65, 6.66, 6.837, 6.808, 6.848, 6.765, 6.675, 6.812, 6.82, 6.753, 6.777, 7.045, 6.72, 6.807, 6.805, 6.747, 6.808, 6.68, 6.962, 6.828, 6.798, 6.835, 6.95, 6.823, 6.682, 6.852, 6.77, 6.613, 6.743, 6.577, 6.628, 6.698), mag.cat = c(3L, 3L, 3L, 3L, 3L, 3L, 3L, 2L, 3L, 3L, 2L, 2L, 1L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 2L, 1L, 1L, 2L, 2L, 1L, 3L, 2L, 2L, 3L, 1L, 2L, 1L, 2L, 2L, 2L, 1L, 2L, 1L, 1L, 1L, 1L, 2L, 1L, 2L, 3L, 2L, 2L, 3L, 3L), names = structure(c(3L, 4L, 9L, 2L, 2L, 11L, 2L, 8L, 10L, 2L, 21L, 29L, 1L, 19L, 1L, 24L, 28L, 1L, 1L, 1L, 23L, 23L, 1L, 27L, 12L, 1L, 15L, 17L, 16L, 10L, 1L, 24L, 1L, 1L, 20L, 14L, 1L, 25L, 1L, 1L, 1L, 1L, 18L, 1L, 22L, 7L, 13L, 26L, 6L, 5L), .Label = c("", "Amen,Assen,Deurze,Ekehaar,Eleveld,Geelbroek,Hooghalen,Marwijksoord,Vredenheim",
"Amen,Assen,Deurze,Ekehaar,Eleveld,Geelbroek,Taarlo,Ubbena",
"Amen,Ekehaar,Eleveld,Geelbroek,Hooghalen", "Annen,Gasteren,Nieuw Annerveen,Oud Annerveen,Schipborg,Zeegse,Zuidlaren","Bedum,Eppenhuizen,Garsthuizen,Huizinge,Kantens,Middelstum,Onderdendam,Rottum,Sint Annen,Startenhuizen,Stedum,Stitswerd,Tinallinge,Toornwerd,Uithuizen,Usquert,Warffum,Westeremden,Westerwijtwerd,Zandeweer",
"Bedum,Huizinge,Kantens,Lellens,Middelstum,Onderdendam,Rottum,Sauwerd,Sint Annen,Stedum,Stitswerd,Thesinge,Tinallinge,Toornwerd,Westeremden,Westerwijtwerd,Wetsinge,Winsum",
"Eleveld,Geelbroek", "Emmen", "Emmer-Compascuum,Roswinkel", "Eppenhuizen,Garsthuizen,Huizinge,Kantens,Middelstum,Oldenzijl,Onderdendam,Rottum,Startenhuizen,Stedum,Stitswerd,Toornwerd,Uithuizen,Westeremden,Westerwijtwerd,Zandeweer",
"Eppenhuizen,Garsthuizen,Huizinge,Kantens,Middelstum,Oldenzijl,Rottum,Startenhuizen,Toornwerd,Westeremden,Zandeweer",
"Eppenhuizen,Garsthuizen,Oldenzijl,Startenhuizen,t-Zandt,Westeremden,Zeerijp,Zijldijk",
"Eppenhuizen,Oldenzijl,Startenhuizen,Uithuizen,Zandeweer", "Froombosch,Hellum,Noordbroek,Sappemeer,Schildwolde,Slochteren", "Garrelsweer", "Garrelsweer,Overschild,Ten Post,Winneweer", "Huizinge,Startenhuizen", "Langelo", "Leermens,Oosterwijtwerd", "Loppersum,Winneweer", "Oosteinde,Roodeschool", "Oosterwijtwerd", "Overschild", "Steendam", "Stitswerd", "t-Zandt,Zeerijp", "Westerbroek", "Woudbloem"), class = "factor")), .Names = c("lllocatie", "lat", "lon", "mag.cat", "names"), class = "data.frame", row.names = c(NA, -50L))
我知道要使用lllocatie变量执行此操作：
df2 <- ddply(df1, .(lllocatie), summarise,
n = as.numeric(length(lllocatie)),
lat = round(mean(lat),3),
lon = round(mean(lon),3),
n.1 = as.numeric(length(lllocatie[mag.cat == 1])),
n.2 = as.numeric(length(lllocatie[mag.cat == 2])),
n.3 = as.numeric(length(lllocatie[mag.cat == 3]))
)
但我想用以下内容来概括：
df2 <- ddply(df1, .(unique(unlist(strsplit(as.character(df1$names), ",")))), summarise,
n = #code giving frequency of each unique name,
lat = #code giving mean for each unique name,
lon = #code giving mean for each unique name,
n.1 = #code giving frequency of each unique name for "mag.cat ==1",
n.2 = #code giving frequency of each unique name for "mag.cat ==1",
n.3 = #code giving frequency of each unique name for "mag.cat ==1"
)
我可以使用例如table(unlist(strsplit(as.character(df1$names), ",")))或table(unlist(strsplit(as.character(df1$names), ","))[df1$mag.cat == 1])获取频率，但我无法确定如何在ddply函数中执行此操作。
任何想法如何解决这个问题？
新的dplyr包可以提供任何帮助吗？
答案 0 :(得分：2)
我预先像
那样重塑数据reshapelllocatie <- function(df1) {
tmp <- strsplit(as.character(df1$names), ",")
len <- sapply(tmp, length)
tmp <- cbind.data.frame(name=unlist(tmp), row=rep(1:nrow(df1), times=len))
tmp <- merge(x=tmp, y=df1, by.x="row", by.y="row.names", all.x=TRUE)[-1]
return(tmp)
}
df2 <- ddply(reshapelllocatie(df1), .(name), summarise,
n = as.numeric(length(name)),
lat = round(mean(lat),3),
lon = round(mean(lon),3),
n.1 = as.numeric(length(name[mag.cat == 1])),
n.2 = as.numeric(length(name[mag.cat == 2])),
n.3 = as.numeric(length(name[mag.cat == 3]))
)
df2
# name n lat lon n.1 n.2 n.3
# 1 Amen 6 52.959 6.565 0 0 6
# 2 Annen 1 53.061 6.698 0 0 1
# 3 Assen 5 52.965 6.568 0 0 5
# ...
答案 1 :(得分：2)
使用strsplit函数和data.table包，无需编写单独的函数即可实现相同的功能：
dt2 <- setDT(df1)[, strsplit(as.character(names), ",", fixed=TRUE), by = setdiff(names(df1),"names")
][, .(.N, lat = round(mean(lat),3),
lon = round(mean(lon),3),
n.1 = as.numeric(length(V1[mag.cat == 1])),
n.2 = as.numeric(length(V1[mag.cat == 2])),
n.3 = as.numeric(length(V1[mag.cat == 3]))), by=V1][order(V1)]
结果：
> dt2
V1 N lat lon n.1 n.2 n.3
1: Amen 6 52.959 6.565 0 0 6
2: Annen 1 53.061 6.698 0 0 1
3: Assen 5 52.965 6.568 0 0 5
.....
这样做的缺点是names列已重命名为V1。使用tstrsplit包中的data.table函数，您会得到相同的结果，并且该列保留名称names：
library(data.table) #v1.9.5
dt2 <- setDT(df1)[, lapply(.SD, function(x) unlist(tstrsplit(x, ",", fixed=TRUE))), by=setdiff(names(df1),"names")
][, .(.N, lat = round(mean(lat),3),
lon = round(mean(lon),3),
n.1 = as.numeric(length(names[mag.cat == 1])),
n.2 = as.numeric(length(names[mag.cat == 2])),
n.3 = as.numeric(length(names[mag.cat == 3]))), by=names][order(names)]
这给出了：
> dt2
names N lat lon n.1 n.2 n.3
1: Amen 6 52.959 6.565 0 0 6
2: Annen 1 53.061 6.698 0 0 1
3: Assen 5 52.965 6.568 0 0 5
.....
另一个替代方案是cSplit包中的splitstackshape函数：
library(splitstackshape)
dt3 <- cSplit(df1, sep=",", "names", 'long',
type.convert=TRUE)[, .(.N, lat = round(mean(lat),3),
lon = round(mean(lon),3),
n.1 = as.numeric(length(names[mag.cat == 1])),
n.2 = as.numeric(length(names[mag.cat == 2])),
n.3 = as.numeric(length(names[mag.cat == 3]))),
by=names][order(names)]
第三种选择是dplyr和tidyr：
library(dplyr)
library(tidyr)
df2 <- df %>%
mutate(names = strsplit(as.character(names),",")) %>%
unnest(names) %>%
group_by(names) %>%
summarise(n = as.numeric(length(names)),
lat = round(mean(lat),3),
lon = round(mean(lon),3),
n.1 = as.numeric(length(names[mag.cat == 1])),
n.2 = as.numeric(length(names[mag.cat == 2])),
n.3 = as.numeric(length(names[mag.cat == 3]))) %>%
arrange(names)
这给出了：
> df2
Source: local data table [69 x 7]
names n lat lon n.1 n.2 n.3
1 Amen 1 52.959 6.565 0 0 6
2 Annen 1 53.061 6.698 0 0 1
3 Assen 1 52.965 6.568 0 0 5
.....