从数据帧列'C'分配/连接值，对应于同一对值'A`&`B`到第二个数据帧。R-dplyr

我有两个数据帧

• 第一个（

  df1

  ）是由字符串

  names1

  和

  names2

  及其

  频率组成的频率数据帧
  第二个（
  df2
  ）包含两列
  names1
  和
  names2
  ，其中包含一个或多个对，或者不包含这些对。有时会有不同的顺序
  

我想在第一个dafarame

df1

的新列中指定频率

df1 <- tibble(names1 = c('architecture', 'assessment', 'build'), 
              names2 = c('build', 'data', 'data'),
              frequency = c(36,13,720))

# A tibble: 3 x 3
  names1       names2 frequency
  <chr>        <chr>      <dbl>
1 architecture build         36
2 assessment   data          13
3 build        data          720

对于此结果：

  names1       names2        frequency
  <chr>        <chr>         <dbl>
1 architecture build         36
2 build        architecture  36
3 assessment   data          13
4 assessment   data          13
5 business     strategy      0

注意：我想保留不匹配的行

5 business     strategy      0

---

这里的问题是，名称列的顺序对于联接很重要，所以必须更新数据集并应用一致的顺序

下面是一个

dplyr

解决方案：

library(dplyr)

df1 <- tibble(names1 = c('architecture', 'assessment', 'build'), 
              names2 = c('build', 'data', 'data'),
              frequency = c(36,13,720))

df2 <- tibble(names1 = c('architecture', 'build', 'assessment','assessment', 'business'), 
              names2 = c('build','architecture', 'data', 'data', 'strategy'))

# update df1
df1 = df1 %>% 
  rowwise() %>% 
  mutate(names = paste0(sort(c(names1, names2)), collapse = "_")) %>% 
  select(names, frequency)

# update df2
df2 = df2 %>% 
  rowwise() %>% 
  mutate(names = paste0(sort(c(names1, names2)), collapse = "_"))

# join datasets and update columns
left_join(df2, df1, by="names") %>%
  mutate(frequency = coalesce(frequency, 0)) %>%
  select(-names) %>%
  ungroup()

#   names1       names2       frequency
#   <chr>        <chr>            <dbl>
# 1 architecture build               36
# 2 build        architecture        36
# 3 assessment   data                13
# 4 assessment   data                13
# 5 business     strategy             0

require(dplyr
require(tidyr) 
left_join(df2,df1,by=c("names1","names2")) %>% 
   left_join(df1,by=c(names1="names2",names2="names1")) %>% 
   mutate(frequency=coalesce(frequency.x,frequency.y,0)) %>% 
   select(-frequency.x,-frequency.y)

库（dplyr）
df1%
mutate（names=paste0（sort（c（names1，names2）），collapse=“\u”））%>%
选择（名称、频率）
#更新df2
df2=df2%>%
行（）
mutate（names=paste0（sort（c（names1，names2）），collapse=“\u1”））
#连接数据集并更新列
左联合（df2，df1，by=“name”）%>%
突变（频率=结合（频率，0））%>%
选择（-names）%%>%
解组（）
#名称1名称2频率
#                       
#1建筑建造36
#2构建架构36
#3评估数据13
#4评估数据13
#5商业策略0

这里的问题是，为了加入，名称列的顺序很重要，所以您必须更新数据集并应用一致的顺序

下面是一个

dplyr

解决方案：

library(dplyr)

df1 <- tibble(names1 = c('architecture', 'assessment', 'build'), 
              names2 = c('build', 'data', 'data'),
              frequency = c(36,13,720))

df2 <- tibble(names1 = c('architecture', 'build', 'assessment','assessment', 'business'), 
              names2 = c('build','architecture', 'data', 'data', 'strategy'))

# update df1
df1 = df1 %>% 
  rowwise() %>% 
  mutate(names = paste0(sort(c(names1, names2)), collapse = "_")) %>% 
  select(names, frequency)

# update df2
df2 = df2 %>% 
  rowwise() %>% 
  mutate(names = paste0(sort(c(names1, names2)), collapse = "_"))

# join datasets and update columns
left_join(df2, df1, by="names") %>%
  mutate(frequency = coalesce(frequency, 0)) %>%
  select(-names) %>%
  ungroup()

#   names1       names2       frequency
#   <chr>        <chr>            <dbl>
# 1 architecture build               36
# 2 build        architecture        36
# 3 assessment   data                13
# 4 assessment   data                13
# 5 business     strategy             0

require(dplyr
require(tidyr) 
left_join(df2,df1,by=c("names1","names2")) %>% 
   left_join(df1,by=c(names1="names2",names2="names1")) %>% 
   mutate(frequency=coalesce(frequency.x,frequency.y,0)) %>% 
   select(-frequency.x,-frequency.y)

库（dplyr）
df1%
mutate（names=paste0（sort（c（names1，names2）），collapse=“\u”））%>%
选择（名称、频率）
#更新df2
df2=df2%>%
行（）
mutate（names=paste0（sort（c（names1，names2）），collapse=“\u1”））
#连接数据集并更新列
左联合（df2，df1，by=“name”）%>%
突变（频率=结合（频率，0））%>%
选择（-names）%%>%
解组（）
#名称1名称2频率
#                       
#1建筑建造36
#2构建架构36
#3评估数据13
#4评估数据13
#5商业策略0

一个递归的

tidyr:：left\u join

与一些

dplyr

解决方案：

library(dplyr)

df1 <- tibble(names1 = c('architecture', 'assessment', 'build'), 
              names2 = c('build', 'data', 'data'),
              frequency = c(36,13,720))

df2 <- tibble(names1 = c('architecture', 'build', 'assessment','assessment', 'business'), 
              names2 = c('build','architecture', 'data', 'data', 'strategy'))

# update df1
df1 = df1 %>% 
  rowwise() %>% 
  mutate(names = paste0(sort(c(names1, names2)), collapse = "_")) %>% 
  select(names, frequency)

# update df2
df2 = df2 %>% 
  rowwise() %>% 
  mutate(names = paste0(sort(c(names1, names2)), collapse = "_"))

# join datasets and update columns
left_join(df2, df1, by="names") %>%
  mutate(frequency = coalesce(frequency, 0)) %>%
  select(-names) %>%
  ungroup()

#   names1       names2       frequency
#   <chr>        <chr>            <dbl>
# 1 architecture build               36
# 2 build        architecture        36
# 3 assessment   data                13
# 4 assessment   data                13
# 5 business     strategy             0

require(dplyr
require(tidyr) 
left_join(df2,df1,by=c("names1","names2")) %>% 
   left_join(df1,by=c(names1="names2",names2="names1")) %>% 
   mutate(frequency=coalesce(frequency.x,frequency.y,0)) %>% 
   select(-frequency.x,-frequency.y)

此解决方案保留df2中列的顺序。之所以有mutate和select行，是因为left_join添加了新的列，这些列需要合并回单个频率列（并用0替换NAs），然后删除

结果:

# A tibble: 5 x 3
  names1       names2       frequency
  <chr>        <chr>            <dbl>
1 architecture build               36
2 build        architecture        36
3 assessment   data                13
4 assessment   data                13
5 business     strategy             0

#一个tible:5 x 3
名称1名称2频率
1建筑建造36
2构建架构36
3评估数据13
4评估数据13
5商业策略0

一个递归的

tidyr:：left\u join

与一些

dplyr

解决方案：

library(dplyr)

df1 <- tibble(names1 = c('architecture', 'assessment', 'build'), 
              names2 = c('build', 'data', 'data'),
              frequency = c(36,13,720))

df2 <- tibble(names1 = c('architecture', 'build', 'assessment','assessment', 'business'), 
              names2 = c('build','architecture', 'data', 'data', 'strategy'))

# update df1
df1 = df1 %>% 
  rowwise() %>% 
  mutate(names = paste0(sort(c(names1, names2)), collapse = "_")) %>% 
  select(names, frequency)

# update df2
df2 = df2 %>% 
  rowwise() %>% 
  mutate(names = paste0(sort(c(names1, names2)), collapse = "_"))

# join datasets and update columns
left_join(df2, df1, by="names") %>%
  mutate(frequency = coalesce(frequency, 0)) %>%
  select(-names) %>%
  ungroup()

#   names1       names2       frequency
#   <chr>        <chr>            <dbl>
# 1 architecture build               36
# 2 build        architecture        36
# 3 assessment   data                13
# 4 assessment   data                13
# 5 business     strategy             0

require(dplyr
require(tidyr) 
left_join(df2,df1,by=c("names1","names2")) %>% 
   left_join(df1,by=c(names1="names2",names2="names1")) %>% 
   mutate(frequency=coalesce(frequency.x,frequency.y,0)) %>% 
   select(-frequency.x,-frequency.y)

此解决方案保留df2中列的顺序。之所以有mutate和select行，是因为left_join添加了新的列，这些列需要合并回单个频率列（并用0替换NAs），然后删除

结果:

# A tibble: 5 x 3
  names1       names2       frequency
  <chr>        <chr>            <dbl>
1 architecture build               36
2 build        architecture        36
3 assessment   data                13
4 assessment   data                13
5 business     strategy             0

#一个tible:5 x 3
名称1名称2频率
1建筑建造36
2构建架构36
3评估数据13
4评估数据13
5商业策略0

从

dplyr

中签出

left\u join

和

right\u join

我的问题是有时我有

df1$A==df2$B和&df1$B==df2$A

或

df1$A==df2$B和&df1$B==df2$A

``1架构构建36``签出

left\u join

和

right\u join

from

dplyr

我的问题是有时我有

df1$A==df2$B和&df1$B==df2$A

或

df1$A==df2$B和&df1$B==df2$A

``1体系结构构建36 2构建体系结构36```