2

我想根据字符串变量的唯一值来总结一个数据框。

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))

lllocatie我知道用变量来做到这一点：

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]))
)

但我想总结一下：

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"
)

table(unlist(strsplit(as.character(df1$names), ",")))例如，我可以使用或获得频率table(unlist(strsplit(as.character(df1$names), ","))[df1$mag.cat == 1])，但我无法弄清楚如何在ddply函数内部进行操作。

任何想法如何解决这个问题？新dplyr包有什么帮助吗？

4

2 回答 2

2

我会事先重塑数据，例如

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
# ...
于 2014-02-28T10:42:46.913 回答
2

使用strsplit函数和data.table包，您可以在不编写单独函数的情况下实现相同的功能：

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
.....

这样做的缺点是该names列被重命名为V1. 使用包中的tstrsplit函数data.table，您会得到相同的结果，并且列保留名称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
 .....

另一种选择是包中的cSplit功能splitstackshape：

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)]

第三种选择是结合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
.....
于 2015-08-02T15:35:07.747 回答