R 创建两个新列并删除源列

我有以下示例数据：

df <- data.frame(ID=c("A1","A2","A3","A4","A1","A2","A3","A4"),
                 NUM=c(469,586,394,595,398,203,604,809))

我希望提取NUM列的第一个值并将其放在新列NUM1中，然后在同一ID的第二个NUM值出现时，将该值提取到新列NUM2中。最后，我想删除原始列。除了ID和NUM之外，我拥有的数据集还有更多的变量和列

df1 <- data.frame(ID=c("A1","A2","A3","A4"),NUM1=c(469,586,394,595),NUM2=c(398,203,604,809))

---

这里有一种方法。您需要创建一个COL列作为新列的名称，因此在本例中，我们使用group_by和str_c来创建此列。pivot_wider是扩展函数的更新版本。所有这些功能都来自tidyverse软件包

library(tidyverse)

df1 <- df %>%
  group_by(ID) %>%
  mutate(COL = str_c("NUM", row_number())) %>%
  pivot_wider(names_from = COL, values_from = NUM) %>%
  ungroup()
df1
# # A tibble: 4 x 3
#   ID     NUM1  NUM2
#   <fct> <dbl> <dbl>
# 1 A1      469   398
# 2 A2      586   203
# 3 A3      394   604
# 4 A4      595   809

---

这里有一种方法。您需要创建一个COL列作为新列的名称，因此在本例中，我们使用group_by和str_c来创建此列。pivot_wider是扩展函数的更新版本。所有这些功能都来自tidyverse软件包

library(tidyverse)

df1 <- df %>%
  group_by(ID) %>%
  mutate(COL = str_c("NUM", row_number())) %>%
  pivot_wider(names_from = COL, values_from = NUM) %>%
  ungroup()
df1
# # A tibble: 4 x 3
#   ID     NUM1  NUM2
#   <fct> <dbl> <dbl>
# 1 A1      469   398
# 2 A2      586   203
# 3 A3      394   604
# 4 A4      595   809

---

您可以通过子集来获得每个ID的第一个和第二个值

library(dplyr)

df %>%
  group_by(ID) %>%
  summarise(NUM1 = NUM[1L], 
            NUM2 = NUM[2L])

# A tibble: 4 x 3
#  ID     NUM1  NUM2
#  <fct> <dbl> <dbl>
#1 A1      469   398
#2 A2      586   203
#3 A3      394   604
#4 A4      595   809

---

您可以通过子集来获得每个ID的第一个和第二个值

library(dplyr)

df %>%
  group_by(ID) %>%
  summarise(NUM1 = NUM[1L], 
            NUM2 = NUM[2L])

# A tibble: 4 x 3
#  ID     NUM1  NUM2
#  <fct> <dbl> <dbl>
#1 A1      469   398
#2 A2      586   203
#3 A3      394   604
#4 A4      595   809

---

使用base R，您可以执行以下操作：

 reshape(transform(df,time=cumsum(grepl("1",ID))),idvar = "ID",dir="wide",sep="")

  ID NUM1 NUM2
1 A1  469  398
2 A2  586  203
3 A3  394  604
4 A4  595  809

或者你可以试试：

`colnames<-`(t(unstack(df,NUM~ID)),c("NUM1","NUM2"))
   NUM1 NUM2
A1  469  398
A2  586  203
A3  394  604
A4  595  809

---

使用base R，您可以执行以下操作：

 reshape(transform(df,time=cumsum(grepl("1",ID))),idvar = "ID",dir="wide",sep="")

  ID NUM1 NUM2
1 A1  469  398
2 A2  586  203
3 A3  394  604
4 A4  595  809

或者你可以试试：

`colnames<-`(t(unstack(df,NUM~ID)),c("NUM1","NUM2"))
   NUM1 NUM2
A1  469  398
A2  586  203
A3  394  604
A4  595  809

---

@akrun雄辩的Base R解决方案：

df1 <- aggregate(NUM ~ ID, df, I)

我的基本R解决方案：

#Transform the dataframe: 

df1 <- within(df, {
              count_num_by_id <- ave(NUM, ID, FUN = seq.int);
              NUM2 <- ifelse(count_num_by_id == 2, NUM, 0);
              NUM <- ifelse(count_num_by_id == 1, NUM, 0);
              rm(count_num_by_id)})

# Aggregate the dataframe: 

df1 <- data.frame(aggregate(.~ID, df1, sum))

---

@akrun雄辩的Base R解决方案：

df1 <- aggregate(NUM ~ ID, df, I)

我的基本R解决方案：

#Transform the dataframe: 

df1 <- within(df, {
              count_num_by_id <- ave(NUM, ID, FUN = seq.int);
              NUM2 <- ifelse(count_num_by_id == 2, NUM, 0);
              NUM <- ifelse(count_num_by_id == 1, NUM, 0);
              rm(count_num_by_id)})

# Aggregate the dataframe: 

df1 <- data.frame(aggregate(.~ID, df1, sum))

数据表解决方案

require(data.table)

# Set as a data.table and create a unique row.
setDT(df)[, rid := paste0('NUM', rowid(ID))]

# Cast the data by ID and rid.
df <- dcast(df, ID ~ rid, value.var = 'NUM')

df
#    ID NUM1 NUM2
# 1: A1  469  398
# 2: A2  586  203
# 3: A3  394  604
# 4: A4  595  809

数据表解决方案

require(data.table)

# Set as a data.table and create a unique row.
setDT(df)[, rid := paste0('NUM', rowid(ID))]

# Cast the data by ID and rid.
df <- dcast(df, ID ~ rid, value.var = 'NUM')

df
#    ID NUM1 NUM2
# 1: A1  469  398
# 2: A2  586  203
# 3: A3  394  604
# 4: A4  595  809

下面是一种dcast方法，它直接在公式中调用rowid，还将处理df中的其他列：

注意调用rowid时的prefix=NUM参数

df中的附加列 OP指出他的数据集[…]除了ID和NUM之外还有更多的变量和列

如果每个ID的附加列的值相同，则+。。。将它们添加到输出中：

df2 <- data.frame(
  ID = c("A1", "A2", "A3", "A4", "A1", "A2", "A3", "A4"),
  NUM = c(469, 586, 394, 595, 398, 203, 604, 809),
  other1 = rep(4:1, 2),
  other2 = rep(letters[1:4], 2)
)

df2

下面是一种dcast方法，它直接在公式中调用rowid，还将处理df中的其他列：

注意调用rowid时的prefix=NUM参数

df中的附加列 OP指出他的数据集[…]除了ID和NUM之外还有更多的变量和列

如果每个ID的附加列的值相同，则+。。。将它们添加到输出中：

df2 <- data.frame(
  ID = c("A1", "A2", "A3", "A4", "A1", "A2", "A3", "A4"),
  NUM = c(469, 586, 394, 595, 398, 203, 604, 809),
  other1 = rep(4:1, 2),
  other2 = rep(letters[1:4], 2)
)

df2

---

@akrun是一个极好的解决方案。我已经修改了我上面的解决方案。我已经在上面修改了我的。