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[英]How to add two columns in a data frame to create a new third column, based on sub-strings of column names, in R?
[英]Add third column to data frame based on values in two existing columns
我有一個與此類似的數據框,其中每個州在 4 個季度內都有多個觀察結果。
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))
我現在想添加第三列,其中每個狀態為 qtr 1 和 2 分配一個值,為季度 3 和 4 分配一個不同的值。我希望結果如下所示:
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
我希望模式很清楚。 我試過這個
df$unemp <- ifelse(df$qtr <3 & df$states %in% "AL",4.4,4.1)
但我不知道如何為其添加更多參數。 這僅創建了 unemp 列,但與參數不匹配。
正如評論中所指出的,最好提供模仿數據的可重現示例。 您想要做的是在一些數據操作之后進行連接(將前兩個 qtr 合並到一個類中,最后兩個相同)。
library(dplyr)
df <- data.frame(states=rep(c("AL","AR","FL","GA","LA","MS","NC","OK","SC","TN","TX"), times = 4),
qtr=rep(c(1,2,3,4), times = 11))
df <- df %>% arrange(states, qtr) # pure cosmetics
df <- df %>% mutate(sem=ifelse(qtr <= 2, 1, 2), # merge the first two and the last two
key=paste0(states, "_", sem)) # create a joining key
head(df)
states qtr sem key
1 AL 1 1 AL_1
2 AL 2 1 AL_1
3 AL 3 2 AL_2
4 AL 4 2 AL_2
5 AR 1 1 AR_1
6 AR 2 1 AR_1
# recreate an external source
ext <- df %>% select(states, sem) %>% distinct()
set.seed(123) # for the sake of reproductibility
ext$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
這就是你要找的嗎?
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