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)通过"id"合并2个数据帧,(2)将两个帧中"hello"和"world"列的所有数据合并为1个"hello"列和1个"world" "列,以及(3)具有与最终数据帧还包含所有在2个原始帧的其他列的("曲棍球",'足球’,'足球’,'棒球’).所以,我希望最终的结果如下:
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
我在r很新,所以唯一的代码我试过都在变化merge,我已经试过我发现这里的答案,这是基于一个类似的问题:R:合并同一个变量的副本.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"你好"和"世界")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编写的东西.
有什么想法可以做到吗？对不起,我无法提供我的努力样本,但我真的不知道从哪里开始:
mydata <- merge(df1, df2, by=c("ID"), all = TRUE)
要重现数据框:
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))
A5C*_*2T1 12
这是一种涉及melt数据,合并熔融数据以及使用dcast将其恢复为宽泛形式的方法.我添加了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
没有人发布dplyr解决方案,所以这里有一个简洁的选择dplyr.这种方法是简单地做一个full_join结合了所有的行,然后group和summarise除去多余的缺失单元.
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包创建于2018-07-13 (v0.2.0).
这是data.table使用二进制合并的另一种方法
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
@ ananda-mahto的答案更优雅,但这是我的建议:
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')
这是一种更为tidyr中心的方法,它与当前接受的答案类似.方法只是将数据框堆叠在一起bind_rows(使列名匹配),gather向上堆叠所有非ID列na.rm = TRUE,然后将spread它们退出.对于条件"如果值为NA in"df1"它将具有"df2"中的值(反之亦然)"的情况应该是稳健的"与summarise选项相比,"并不总是成立.
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包创建于2018-07-13 (v0.2.0).
使用tidyverse我们可以使用coalesce.
下面的解决方案都没有构建额外的行，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
listtranspose它，因此根中的每个新项目都有输出列的名称。的默认行为transpose是将第一个参数作为模板，因此不幸的是，我们必须明确获取所有参数。compact这些项目，因为它们的长度都是 2，但其中一个是NULL当给定的列在一侧丢失时。coalesce那些，这基本上意味着在NA并排放置参数时返回您找到的第一个非。如果在第二行重复df1和df2出现问题，请改用以下内容：
transpose(invoke(union, setNames(map(., names), c("x","y"))))
解决方案2
相同的理念，但这次我们循环名称：
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更喜欢的人提供：
union(names(df1), names(df2)) %>%
set_names %>%
map_dfc(~ list(df1[[.x]], df2[[.x]]) %>%
compact %>%
invoke(coalesce, .))
说明
set_names给出与其值相同的字符向量名称，因此map_dfc可以正确命名输出的列。df1[[.x]]NULL当.x不是 的列时将返回df1，我们利用这一点。df1并且df2每次被提及 2 次，我想不出任何解决方法。解决方案 1 在这些方面更清晰，所以我推荐它。