Question

我有一个与此类似的数据框，其中每个州在4个季度中都有一些观测结果。

df <- data.frame(states=rep(c("AL","AR","FL","GA","LA","MS","NC","OK","SC","TN","TX"), times = 4),
qtr=rep(c(1,2,3,4), times = 11))

我现在想添加第三列，在每个状态中为qtr 1和2分配一个值，为四分之三和4分配一个不同的值。我希望结果看起来像这样：

state qtr unemp
AL    1   4.4
AL    2   4.4
AL    3   4.1
AL    4   4.1 
AR    1   3.7 
AR    2   3.7 
AR    3   3.9
AR    4   3.9

我希望模式清楚。我尝试过

df$unemp <- ifelse(df$qtr <3 & df$states %in% "AL",4.4,4.1)

但是我不知道如何向它添加更多参数。这仅创建unemp列，但不与参数匹配。

Answer 1

正如评论中指出的那样，最好提供可复制的示例来模仿您的数据。您要做的是在进行一些数据操作后将 join （将前两个qtr合并到一个类中，最后两个相同）。

library(dplyr)
df <- data.frame(states=rep(c("AL","AR","FL","GA","LA","MS","NC","OK","SC","TN","TX"), times = 4),
                 qtr=rep(c(1,2,3,4), times = 11))

df <- df %>% arrange(states, qtr) # pure cosmetics
df <- df %>% mutate(sem=ifelse(qtr <= 2, 1, 2),    # merge the first two and the last two
                    key=paste0(states, "_", sem))  # create a joining key


head(df)
states qtr sem  key
1     AL   1   1 AL_1
2     AL   2   1 AL_1
3     AL   3   2 AL_2
4     AL   4   2 AL_2
5     AR   1   1 AR_1
6     AR   2   1 AR_1

# recreate an external source
ext <- df %>% select(states, sem) %>% distinct()

set.seed(123) # for the sake of reproductibility
ext$unemp <- runif(nrow(ext)/2) # simulate some unemp rates
# you probably have something that looks like this:
head(ext)
states sem     unemp
1     AL   1 0.2875775
2     AL   2 0.7883051
3     AR   1 0.4089769
4     AR   2 0.8830174
5     FL   1 0.9404673
6     FL   2 0.0455565

# recreate a key column
ext <- mutate(ext, key=paste0(states, "_", sem))

# have a look at it
head(ext)
states sem     unemp  key
1     AL   1 0.2875775 AL_1
2     AL   2 0.7883051 AL_2
3     AR   1 0.4089769 AR_1
4     AR   2 0.8830174 AR_2
5     FL   1 0.9404673 FL_1
6     FL   2 0.0455565 FL_2

# left join and drop redundant columns
df2 <- left_join(df, ext, "key") %>% 
  transmute(states=states.x, qtr, unemp)

head(df2)
states qtr     unemp
1     AL   1 0.2875775
2     AL   2 0.2875775
3     AL   3 0.7883051
4     AL   4 0.7883051
5     AR   1 0.4089769
6     AR   2 0.4089769

您正在寻找什么吗？

根据两个现有列中的值将第三列添加到数据框

1 个答案: