我有两个数据框,其中一些列具有相同的名称,另一些具有不同的名称。数据框看起来像这样:
df1
ID hello world hockey soccer
1 1 NA NA 7 4
2 2 NA NA 2 5
3 3 10 8 8 23
4 4 4 17 5 12
5 5 NA NA 3 43
df2
ID hello world football baseball
1 1 2 3 43 6
2 2 5 1 24 32
3 3 NA NA 2 23
4 4 NA NA 5 15
5 5 9 7 12 23
如您所见,在2个共享列(“hello”和“world”)中,某些数据位于其中一个数据框中,其余数据位于另一个数据框中。
我要做的是(1)将2个数据帧合并为“id”,(2)将两个帧中“hello”和“world”列的所有数据合并为1个“hello”列, 1“世界”列,以及(3)最终数据框还包含2个原始帧中的所有其他列(“曲棍球”,“足球”,“足球”,“棒球”)。所以,我希望最终结果如下:
ID hello world hockey soccer football baseball
1 1 2 3 7 4 43 6
2 2 5 3 2 5 24 32
3 3 10 8 8 23 2 23
4 4 4 17 5 12 5 15
5 5 9 7 3 43 12 23
我是R的新手,所以我尝试的唯一代码是merge
的变体,我尝试了我在这里找到的答案,该答案基于类似的问题:R: merging copies of the same variable 。但是,我的数据集实际上比我在这里显示的要大得多(大约有20个匹配的列(如“hello”和“world”)和100个不匹配的列(如“曲棍球”和“足球”))所以我正在寻找一些不需要我手动全部写出来的东西。
知道是否可以这样做?对不起,我不能提供我的努力样本,但我真的不知道从哪里开始:
mydata <- merge(df1, df2, by=c("ID"), all = TRUE)
重现数据框:
df1 <- structure(list(ID = c(1L, 2L, 3L, 4L, 5L), hellow = c(2, 5, NA, NA, 9),
world = c(3, 1, NA, NA, 7), football = c(43, 24, 2, 5, 12),
baseball = c(6, 32, 23, 15, 23)), .Names = c("ID", "hello", "world",
"football", "baseball"), class = "data.frame", row.names = c(NA, -5L))
df2 <- structure(list(ID = c(1L, 2L, 3L, 4L, 5L), hellow = c(NA, NA, 10, 4, NA),
world = c(NA, NA, 8, 17, NA), hockey = c(7, 2, 8, 5, 3),
soccer = c(4, 5, 23, 12, 43)), .Names = c("ID", "hello", "world", "hockey",
"soccer"), class = "data.frame", row.names = c(NA, -5L))
答案 0 :(得分:12)
这是一种涉及melt
数据,合并熔融数据,并使用dcast
将其恢复为宽泛形式的方法。我已添加评论以帮助了解正在发生的事情。
## Required packages
library(data.table)
library(reshape2)
dcast.data.table(
merge(
## melt the first data.frame and set the key as ID and variable
setkey(melt(as.data.table(df1), id.vars = "ID"), ID, variable),
## melt the second data.frame
melt(as.data.table(df2), id.vars = "ID"),
## you'll have 2 value columns...
all = TRUE)[, value := ifelse(
## ... combine them into 1 with ifelse
is.na(value.x), value.y, value.x)],
## This is your reshaping formula
ID ~ variable, value.var = "value")
# ID hello world football baseball hockey soccer
# 1: 1 2 3 43 6 7 4
# 2: 2 5 1 24 32 2 5
# 3: 3 10 8 2 23 8 23
# 4: 4 4 17 5 15 5 12
# 5: 5 9 7 12 23 3 43
答案 1 :(得分:8)
没有人发布过dplyr
解决方案,因此这是dplyr
中的简洁选项。该方法只是要做一个full_join
合并所有行,然后group
和summarise
删除多余的丢失单元格。
library(tidyverse)
df1 <- structure(list(ID = 1:5, hello = c(NA, NA, 10L, 4L, NA), world = c(NA, NA, 8L, 17L, NA), hockey = c(7L, 2L, 8L, 5L, 3L), soccer = c(4L, 5L, 23L, 12L, 43L)), row.names = c(NA, -5L), class = c("tbl_df", "tbl", "data.frame"), spec = structure(list(cols = list(ID = structure(list(), class = c("collector_integer", "collector")), hello = structure(list(), class = c("collector_integer", "collector")), world = structure(list(), class = c("collector_integer", "collector")), hockey = structure(list(), class = c("collector_integer", "collector")), soccer = structure(list(), class = c("collector_integer", "collector"))), default = structure(list(), class = c("collector_guess", "collector"))), class = "col_spec"))
df2 <- structure(list(ID = 1:5, hello = c(2L, 5L, NA, NA, 9L), world = c(3L, 1L, NA, NA, 7L), football = c(43L, 24L, 2L, 5L, 12L), baseball = c(6L, 32L, 23L, 15L, 2L)), row.names = c(NA, -5L), class = c("tbl_df", "tbl", "data.frame"), spec = structure(list(cols = list(ID = structure(list(), class = c("collector_integer", "collector")), hello = structure(list(), class = c("collector_integer", "collector")), world = structure(list(), class = c("collector_integer", "collector")), football = structure(list(), class = c("collector_integer", "collector")), baseball = structure(list(), class = c("collector_integer", "collector"))), default = structure(list(), class = c("collector_guess", "collector"))), class = "col_spec"))
df1 %>%
full_join(df2, by = intersect(colnames(df1), colnames(df2))) %>%
group_by(ID) %>%
summarize_all(na.omit)
#> # A tibble: 5 x 7
#> ID hello world hockey soccer football baseball
#> <int> <int> <int> <int> <int> <int> <int>
#> 1 1 2 3 7 4 43 6
#> 2 2 5 1 2 5 24 32
#> 3 3 10 8 8 23 2 23
#> 4 4 4 17 5 12 5 15
#> 5 5 9 7 3 43 12 2
由reprex package(v0.2.0)于2018-07-13创建。
答案 2 :(得分:6)
这是使用二进制合并的另一种data.table
方法
library(data.table)
setkey(setDT(df1), ID) ; setkey(setDT(df2), ID) # Converting to data.table objects and setting keys
df1 <- df1[df2][, `:=`(i.hello = NULL, i.world = NULL)] # Full left join
df1[df2[complete.cases(df2)], `:=`(hello = i.hello, world = i.world)][] # Joining only on non-missing values
# ID hello world football baseball hockey soccer
# 1: 1 2 3 43 6 7 4
# 2: 2 5 1 24 32 2 5
# 3: 3 10 8 2 23 8 23
# 4: 4 4 17 5 15 5 12
# 5: 5 9 7 12 23 3 43
答案 3 :(得分:5)
library(reshape2)
df1=melt(df1,id='ID',na.rm=TRUE)
df2=melt(df2,id='ID',na.rm=TRUE)
DF=rbind(df1,df2)
# Not needeed, added na.rm=TRUE based on @ananda-mahto's valid comment
# DF<-DF[!is.na(DF$value),]
dcast(DF,ID~variable,value.var='value')
答案 4 :(得分:5)
这是一种更tidyr
为中心的方法,它的作用类似于当前接受的答案。该方法仅是使用bind_rows
(与列名匹配)将数据帧彼此堆叠,将gather
上的所有非ID
列与na.rm = TRUE
堆叠在一起,然后然后spread
将其退出。与summarise
选项相比,条件“如果该值在“ df1”中为NA,则该值在“ df2”中(反之亦然)”并不总是成立的情况应该是健壮的。
library(tidyverse)
df1 <- structure(list(ID = 1:5, hello = c(NA, NA, 10L, 4L, NA), world = c(NA, NA, 8L, 17L, NA), hockey = c(7L, 2L, 8L, 5L, 3L), soccer = c(4L, 5L, 23L, 12L, 43L)), row.names = c(NA, -5L), class = c("tbl_df", "tbl", "data.frame"), spec = structure(list(cols = list(ID = structure(list(), class = c("collector_integer", "collector")), hello = structure(list(), class = c("collector_integer", "collector")), world = structure(list(), class = c("collector_integer", "collector")), hockey = structure(list(), class = c("collector_integer", "collector")), soccer = structure(list(), class = c("collector_integer", "collector"))), default = structure(list(), class = c("collector_guess", "collector"))), class = "col_spec"))
df2 <- structure(list(ID = 1:5, hello = c(2L, 5L, NA, NA, 9L), world = c(3L, 1L, NA, NA, 7L), football = c(43L, 24L, 2L, 5L, 12L), baseball = c(6L, 32L, 23L, 15L, 2L)), row.names = c(NA, -5L), class = c("tbl_df", "tbl", "data.frame"), spec = structure(list(cols = list(ID = structure(list(), class = c("collector_integer", "collector")), hello = structure(list(), class = c("collector_integer", "collector")), world = structure(list(), class = c("collector_integer", "collector")), football = structure(list(), class = c("collector_integer", "collector")), baseball = structure(list(), class = c("collector_integer", "collector"))), default = structure(list(), class = c("collector_guess", "collector"))), class = "col_spec"))
df1 %>%
bind_rows(df2) %>%
gather(variable, value, -ID, na.rm = TRUE) %>%
spread(variable, value)
#> # A tibble: 5 x 7
#> ID baseball football hello hockey soccer world
#> <int> <int> <int> <int> <int> <int> <int>
#> 1 1 6 43 2 7 4 3
#> 2 2 32 24 5 2 5 1
#> 3 3 23 2 10 8 23 8
#> 4 4 15 5 4 5 12 17
#> 5 5 2 12 9 3 43 7
由reprex package(v0.2.0)于2018-07-13创建。
答案 5 :(得分:4)
使用tidyverse
,我们可以使用coalesce
。
以下任何一种解决方案都无法建立额外的行,数据在整个链中或多或少保持相同的大小和相似的形状。
解决方案1
list(df1,df2) %>%
transpose(union(names(df1),names(df2))) %>%
map_dfc(. %>% compact %>% invoke(coalesce,.))
# # A tibble: 5 x 7
# ID hello world football baseball hockey soccer
# <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 1 2 3 43 6 7 4
# 2 2 5 1 24 32 2 5
# 3 3 10 8 2 23 8 23
# 4 4 4 17 5 15 5 12
# 5 5 9 7 12 23 3 43
说明
list
transpose
,因此位于根目录的每个新项目都具有输出列的名称。 transpose
的默认行为是将第一个参数作为模板,因此很遗憾,我们必须明确地获取所有参数。compact
这些项目都是长度2,但是当给定的一列在一侧缺失时,其中一项是NULL
。coalesce
这些基本上意味着在并排放置参数时返回找到的第一个非NA
。如果在第二行重复df1
和df2
是一个问题,请改用以下内容:
transpose(invoke(union, setNames(map(., names), c("x","y"))))
解决方案2
相同的哲学,但是这次我们循环使用名称:
map_dfc(set_names(union(names(df1), names(df2))),
~ invoke(coalesce, compact(list(df1[[.x]], df2[[.x]]))))
# # A tibble: 5 x 7
# ID hello world football baseball hockey soccer
# <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 1 2 3 43 6 7 4
# 2 2 5 1 24 32 2 5
# 3 3 10 8 2 23 8 23
# 4 4 4 17 5 15 5 12
# 5 5 9 7 12 23 3 43
这里曾经为那些可能会喜欢的人而设:
union(names(df1), names(df2)) %>%
set_names %>%
map_dfc(~ list(df1[[.x]], df2[[.x]]) %>%
compact %>%
invoke(coalesce, .))
说明
set_names
赋予字符向量名称与其值相同的名称,因此map_dfc
可以正确命名输出的列。df1[[.x]]
不是NULL
的列时,.x
将返回df1
,我们利用这一点。df1
和df2
分别被提及2次,我想不起来。在这些方面,解决方案1较为干净,因此我建议使用。
答案 6 :(得分:0)
我们可以使用我的软件包 safejoin ,进行左连接并使用dplyr::coalesce
# # devtools::install_github("moodymudskipper/safejoin")
library(safejoin)
library(dplyr)
safe_left_join(df1, df2, by = "ID", conflict = coalesce)
# # A tibble: 5 x 7
# ID hello world hockey soccer football baseball
# <int> <int> <int> <int> <int> <int> <int>
# 1 1 2 3 7 4 43 6
# 2 2 5 1 2 5 24 32
# 3 3 10 8 8 23 2 23
# 4 4 4 17 5 12 5 15
# 5 5 9 7 3 43 12 2