所以我有一个包含49行和109个特征的数据集，其中数据被格式化，以便每个条目都有一个均值和sd值。这是一个示例：
> head(score_data[,1:4])
# A tibble: 6 x 4
Variable Overall `18 to 29` `30 to 39`
<chr> <chr> <chr> <chr>
1 ts.tsmart_partisan_score (mean (sd)) 94.01 (9.73) 92.56 (10.82) 94.14 (9.55)
2 ts.tsmart_presidential_general_turnout_score (mean (sd)) 66.23 (24.38) 51.56 (20.02) 58.44 (24.36)
3 ts.tsmart_midterm_general_turnout_score (mean (sd)) 50.29 (29.05) 31.09 (18.81) 34.82 (22.15)
4 ts.tsmart_offyear_general_turnout_score (mean (sd)) 20.71 (15.08) 25.38 (17.36) 18.84 (14.35)
5 ts.tsmart_presidential_primary_turnout_score (mean (sd)) 48.34 (28.12) 38.26 (22.26) 36.19 (22.72)
6 ts.tsmart_non_presidential_primary_turnout_score (mean (sd)) 40.21 (29.00) 27.03 (20.14) 23.52 (19.32)
我希望从数据中提取数据集中所有109列的平均值。由于这些特征是字符，我知道我可以使用单独的命令根据第一个括号的索引将数据拆分成两列，如下所示：
data <- data %>% separate(PrecinctName, into = c("Precinct", "PrecinctCode"), sep = 5)
但是，我想为整个数据集中的每个功能执行此操作，并且使用上述方法将非常耗时，并且很痛苦。有没有人有更优雅的解决方案？我并不特别关注保留sd数据，因此该方法不必在其函数中包含该数据。
根据要求，这是替代输出：
> dput( head(score_data[,1:4]))
structure(list(Variable = c("ts.tsmart_partisan_score (mean (sd))",
"ts.tsmart_presidential_general_turnout_score (mean (sd))", "ts.tsmart_midterm_general_turnout_score (mean (sd))",
"ts.tsmart_offyear_general_turnout_score (mean (sd))", "ts.tsmart_presidential_primary_turnout_score (mean (sd))",
"ts.tsmart_non_presidential_primary_turnout_score (mean (sd))"
), Overall = c("94.01 (9.73)", "66.23 (24.38)", "50.29 (29.05)",
"20.71 (15.08)", "48.34 (28.12)", "40.21 (29.00)"), `18 to 29` = c("92.56 (10.82)",
"51.56 (20.02)", "31.09 (18.81)", "25.38 (17.36)", "38.26 (22.26)",
"27.03 (20.14)"), `30 to 39` = c("94.14 (9.55)", "58.44 (24.36)",
"34.82 (22.15)", "18.84 (14.35)", "36.19 (22.72)", "23.52 (19.32)"
)), .Names = c("Variable", "Overall", "18 to 29", "30 to 39"), row.names = c(NA,
-6L), class = c("tbl_df", "tbl", "data.frame"))
答案 0 :(得分：1)
read.table(stringsAsFactors = F,text=gsub("\\(.*?\\)|\\)","",do.call(paste,dat)))
V1 V3 V4 V5
1 ts.tsmart_partisan_score 94.01 92.56 94.14
2 ts.tsmart_presidential_general_turnout_score 66.23 51.56 58.44
3 ts.tsmart_midterm_general_turnout_score 50.29 31.09 34.82
4 ts.tsmart_offyear_general_turnout_score 20.71 25.38 18.84
5 ts.tsmart_presidential_primary_turnout_score 48.34 38.26 36.19
6 ts.tsmart_non_presidential_primary_turnout_score 40.21 27.03 23.52
答案 1 :(得分：0)
这样的东西？
means <- sapply(score_data[, -1], function(x) as.numeric(substr(x, 1,
regexpr(" ", x) - 1)))
means
# Overall 18 to 29 30 to 39
# [1,] 94.01 92.56 94.14
# [2,] 66.23 51.56 58.44
# [3,] 50.29 31.09 34.82
# [4,] 20.71 25.38 18.84
# [5,] 48.34 38.26 36.19
# [6,] 40.21 27.03 23.52
答案 2 :(得分：0)
一个简单的正则表达式应该这样做：
for (i in names(score_data)[-(1)]) {
score_data[[i]] <- as.numeric(gsub( " .*$", "", score_data[[i]] ))
}
答案 3 :(得分：0)
您可以使用gsub（）和正则表达式删除括号内的任何字符，如下所示：
test <- score_data %>% mutate_at(vars(-Variable),funs(gsub("\\([^\\)]+\\)", "", ., perl = T)))
Variable Overall X18.to.29 X30.to.39
1 ts.tsmart_partisan_score (mean (sd)) 94.01 92.56 94.14
2 ts.tsmart_presidential_general_turnout_score (mean (sd)) 66.23 51.56 58.44
3 ts.tsmart_midterm_general_turnout_score (mean (sd)) 50.29 31.09 34.82
4 ts.tsmart_offyear_general_turnout_score (mean (sd)) 20.71 25.38 18.84
5 ts.tsmart_presidential_primary_turnout_score (mean (sd)) 48.34 38.26 36.19
6 ts.tsmart_non_presidential_primary_turnout_score (mean (sd)) 40.21 27.03 23.52