如何合并这些数据框并获取列的意思？答案

【问题标题】：How do I merge these dataframes and take the column means?如何合并这些数据框并获取列的意思？
【发布时间】：2016-09-30 05:16:19
【问题描述】：

我已经合并了这两个数据框，我想取列的平均值并将它们组合成一个数据框。这是我合并的数据框：

           ARN     ADOPTION.x    EUTHANASIA.x     OTHER.x           ADOPTION.y     EUTHANASIA.y     OTHER.y
99645 A1428364 0.998701572418 0.0000001962661 0.001298154    0.990897715091705 0.00000014520751 0.009102128
99646 A1428368 0.000122911530 0.9927836656570 0.007093379  0.00270643248222768 0.00152872898616 0.995764852
99647 A1428369 0.978477597237 0.0000004987147 0.021521868    0.933309495449066 0.00000007226861 0.066690445
99648 A1428370 0.004124862142 0.8476260900497 0.148249015 0.000234879378695041 0.87564605474472 0.124119066
99649 A4081681 0.464215517044 0.0024338855874 0.533350587    0.699311196804047 0.04458642378449 0.256102324
99650 A4087524 0.000006879755 0.0021564548369 0.997836649 0.000240802211919799 0.00144882441964 0.998310328

例如，我想对采用.x + 采用.y 的每一列取平均值，然后将列设为“采用”。然后对每一列重复。我该怎么做呢？

【问题讨论】：

您真的要合并它们吗？还是使用 rbind(df1,df2) 更好？

标签： r dataframe merge mean

【解决方案1】：

我们可以对以.x 和.y 结尾的列进行子集化并取平均值

d1 <- (df1[grep("\\.x$", names(df1))] + df1[grep("\\.y$", names(df1))])/2
colnames(d1) <- c("ADOPTION",   "EUTHANASIA", "OTHER")
d2 <- cbind(df1["ARN"], d1)
d2
#        ARN    ADOPTION      EUTHANASIA       OTHER
#99645 A1428364 0.994799644 0.0000001707368 0.005200141
#99646 A1428368 0.001414672 0.4971561973216 0.501429116
#99647 A1428369 0.955893546 0.0000002854917 0.044106157
#99648 A1428370 0.002179871 0.8616360723972 0.136184040
#99649 A4081681 0.581763357 0.0235101546859 0.394726456
#99650 A4087524 0.000123841 0.0018026396283 0.998073488

或者另一种选择是将melt 转换为长格式，然后使用dcast 转换为“宽”

library(data.table)
dM <- melt(setDT(df1), measure = patterns("\\.x$", "\\.y$"), value.name = 
       c("x", "y"))[, variable := c("ADOPTION",   "EUTHANASIA", "OTHER")[variable]
       ][, .(ARN, variable, value = (x+y)/2)]
dcast(dM, ARN~variable, value.var="value")
#      ARN    ADOPTION      EUTHANASIA       OTHER
#1: A1428364 0.994799644 0.0000001707368 0.005200141
#2: A1428368 0.001414672 0.4971561973216 0.501429116
#3: A1428369 0.955893546 0.0000002854917 0.044106157
#4: A1428370 0.002179871 0.8616360723972 0.136184040
#5: A4081681 0.581763357 0.0235101546859 0.394726456
#6: A4087524 0.000123841 0.0018026396283 0.998073488

【讨论】：

【解决方案2】：

@akrun 建议的长格式 + 聚合方法的变体：

datlong <- reshape(dat, idvar="ARN", sep=".", direction="long", varying=-1)
nms <- names(datlong)[-c(1,2)]
datlong$seq <- with(datlong, ave(time,time,FUN=seq_along) )
aggregate(datlong[nms], datlong[c("ARN","seq")], FUN=mean)[-2]

#       ARN    ADOPTION   EUTHANASIA       OTHER
#1 A1428364 0.994799644 1.707368e-07 0.005200141
#2 A1428368 0.001414672 4.971562e-01 0.501429116
#3 A1428369 0.955893546 2.854917e-07 0.044106157
#4 A1428370 0.002179871 8.616361e-01 0.136184040
#5 A4081681 0.581763357 2.351015e-02 0.394726456
#6 A4087524 0.000123841 1.802640e-03 0.998073488

【讨论】：

【解决方案3】：

Hadleyverse 版本：

library(dplyr)
library(tidyr)

       # melt data into long form
df %>% gather(var, val, -ARN) %>% 
    # separate suffixes into new column
    separate(var, c('var', 'origin')) %>% 
    # spread back to wide form
    spread(var, val) %>% 
    # set grouping for aggregation
    group_by(ARN) %>% 
    # aggregate each variable within each group
    summarise_each(funs(mean), -origin)

# Source: local data frame [6 x 4]
# 
#        ARN    ADOPTION   EUTHANASIA       OTHER
#     (fctr)       (dbl)        (dbl)       (dbl)
# 1 A1428364 0.994799644 1.707368e-07 0.005200141
# 2 A1428368 0.001414672 4.971562e-01 0.501429116
# 3 A1428369 0.955893546 2.854917e-07 0.044106157
# 4 A1428370 0.002179871 8.616361e-01 0.136184040
# 5 A4081681 0.581763357 2.351015e-02 0.394726456
# 6 A4087524 0.000123841 1.802640e-03 0.998073488

请注意，与合并相比，按 ID 列、行绑定和聚合进行分组会更容易，即

bind_rows(df1, df2) %>% 
    group_by(ARN) %>% 
    summarise_each(funs(mean))

或在基础R中

aggregate(. ~ ARN, rbind(df1, df2), mean)

或data.table

library(data.table)

setDT(rbind(df1, df2))[, lapply(.SD, mean), by = ARN]

数据

df <- structure(list(ARN = structure(1:6, .Label = c("A1428364", "A1428368", 
    "A1428369", "A1428370", "A4081681", "A4087524"), class = "factor"), 
        ADOPTION.x = c(0.998701572418, 0.00012291153, 0.978477597237, 
        0.004124862142, 0.464215517044, 6.879755e-06), EUTHANASIA.x = c(1.962661e-07, 
        0.992783665657, 4.987147e-07, 0.8476260900497, 0.0024338855874, 
        0.0021564548369), OTHER.x = c(0.001298154, 0.007093379, 0.021521868, 
        0.148249015, 0.533350587, 0.997836649), ADOPTION.y = c(0.990897715091705, 
        0.00270643248222768, 0.933309495449066, 0.000234879378695041, 
        0.699311196804047, 0.000240802211919799), EUTHANASIA.y = c(1.4520751e-07, 
        0.00152872898616, 7.226861e-08, 0.87564605474472, 0.04458642378449, 
        0.00144882441964), OTHER.y = c(0.009102128, 0.995764852, 
        0.066690445, 0.124119066, 0.256102324, 0.998310328)), .Names = c("ARN", 
    "ADOPTION.x", "EUTHANASIA.x", "OTHER.x", "ADOPTION.y", "EUTHANASIA.y", 
    "OTHER.y"), row.names = c(NA, -6L), class = "data.frame")

df1 <- structure(list(ARN = structure(1:6, .Label = c("A1428364", "A1428368", 
    "A1428369", "A1428370", "A4081681", "A4087524"), class = "factor"), 
        ADOPTION = c(0.998701572418, 0.00012291153, 0.978477597237, 
        0.004124862142, 0.464215517044, 6.879755e-06), EUTHANASIA = c(1.962661e-07, 
        0.992783665657, 4.987147e-07, 0.8476260900497, 0.0024338855874, 
        0.0021564548369), OTHER = c(0.001298154, 0.007093379, 0.021521868, 
        0.148249015, 0.533350587, 0.997836649)), row.names = c(NA, 
    -6L), class = "data.frame", .Names = c("ARN", "ADOPTION", "EUTHANASIA", 
    "OTHER"))

df2 <- structure(list(ARN = structure(1:6, .Label = c("A1428364", "A1428368", 
    "A1428369", "A1428370", "A4081681", "A4087524"), class = "factor"), 
        ADOPTION = c(0.990897715091705, 0.00270643248222768, 0.933309495449066, 
        0.000234879378695041, 0.699311196804047, 0.000240802211919799
        ), EUTHANASIA = c(1.4520751e-07, 0.00152872898616, 7.226861e-08, 
        0.87564605474472, 0.04458642378449, 0.00144882441964), OTHER = c(0.009102128, 
        0.995764852, 0.066690445, 0.124119066, 0.256102324, 0.998310328
        )), row.names = c(NA, -6L), class = "data.frame", .Names = c("ARN", 
    "ADOPTION", "EUTHANASIA", "OTHER"))

【讨论】：