用R中不同的模式替换相同字符串的多个匹配

Question

我有一组数据帧，在每一帧中，一个列中都会出现多个相同的字符串，但它们确实反映了不同的观察结果。

library(dplyr)
govs <- c("Government", "Federal", "General government", "Government enterprises", "State and local", 
          "General government", "Government enterprises")
df <- data.frame("gov_levels" = govs, revenue = rnorm(7, mean = 1000, sd = 50))
df

我想用不同的模式替换(或串联)每个事件，这样它们就变得不同了。此代码将返回所需的输出，

    df %>% 
    mutate(gov_levels = stri_replace_first_fixed(str = gov_levels, pattern = "General government", 
                                    replacement = c("Federal general government", 
                                                    "State and local general government")))

但是，取决于"General government“是在偶数行还是奇数行，这是不一致的，正如我在变异前删除第一行时所说明的那样：

    df %>%
    filter(gov_levels != "Government") %>%
    mutate(gov_levels = stri_replace_first_fixed(str = gov_levels, pattern = "General government", 
                                           replacement = c("Federal general government", 
                                                           "State and local general government"))) 

这会导致替换以错误的顺序发生。我正在寻找一种方法来始终如一地应用这个方法，这样它就不会依赖于要替换的字符串的行位置。也就是说，第一场比赛将永远由联邦政府取代，第二场比赛将永远由州和地方政府取代。

根据GEORGE的数据帧的答复列表进行更新，但有一些不一致之处：

govs <- c("Government", "Federal", "General government", "Government enterprises", "State and local", 
          "General government", "Government enterprises", NA, NA)

df1 <- data.frame("col_1" = "col1data", "gov_levels" = govs, revenue = c(rnorm(7, mean = 100, sd = 50), NA, NA), stringsAsFactors = FALSE)
df2 <- data.frame("col_1" = "col1data", "gov_types" = govs, revenue = c(rnorm(7, mean = 100, sd = 50), NA, NA), stringsAsFactors = FALSE)

df2 <- df2 %>%
       filter(gov_types != "Government")

df_list <- list(df1, df2)

用lapply实现George的解决方案来处理我提到的其他问题--我很好奇是否有更好的方法来解决这个问题？

newlevels_gen <- c("Federal general government", "State and local general government")

df_list <- lapply(df_list, 
                  function(x) {x[, 2] <- as.factor(x[, 2]) 
                               return(x)
                               }
                  )

df_list <- lapply(df_list, function(x) {levels(x[,2]) <- c(levels(x[,2]), newlevels_gen)
                                        return(x)
                                        }
                  )

df_list_clean_a <- lapply(df_list, function(x) {x[,2][!is.na(x[,2]) & x[,2] == "General government"] <- newlevels_gen 
                                               return(x)
                                               } 
                         )

Answer 1

这能达到你的目标吗？只需将满足条件的元素替换为所需的字符串向量即可。但是首先，您需要为您的因子添加允许的级别，否则您将得到一个错误。

# First define a string containing the new levels for the 'gov_levels' factor
newlevels <- c("Federal general government", "State and local general government")

# Then add them so that they are allowed as factor levels
levels(df$gov_levels) <- c(levels(df$gov_levels), newlevels)

# Now just replace the values where 'gov_levels' is "General government" with the new string
# They will naturally be assigned in the same order they occur in the dataset
df$gov_levels[df$gov_levels=="General government"] <- newlevels

当然，只有在有两次发生的情况下，这才能奏效，但我想从你的问题上看，情况是这样的吗？如果有两个或更少的，那么我们可以调整最后一行来计算所需出现的次数。

编辑

如果向量gov_levels是一个字符向量，而不是一个因素，那么您不需要担心级别，您可以这样做：

df$gov_levels[df$gov_levels=="General government"] <- 
      c("Federal general government", "State and local general")