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如何在单个函数中完成几个字符向量格式化步骤?

character vector regex r

Conner M. · 技术社区 · 8 年前

已编辑

我有一个简单的列名列表,我想更改其格式,最好是通过编程。以下是列表示例:

    vars_list <- c("tBodyAcc.mean...X", "tBodyAcc.mean...Y", "tBodyAcc.mean...Z",
    "tBodyAcc.std...X", "tBodyAcc.std...Y", "tBodyAcc.std...Z", 
    "tGravityAcc.mean...X", "tGravityAcc.mean...Y", "tGravityAcc.mean...Z",
    "tGravityAcc.std...X", "tGravityAcc.std...Y", "tGravityAcc.std...Z",
    "fBodyAcc.mean...X", "fBodyAcc.mean...Y", "fBodyAcc.mean...Z", 
    "fBodyAcc.std...X", "fBodyAcc.std...Y", "fBodyAcc.std...Z",
    "fBodyAccJerk.mean...X", "fBodyAccJerk.mean...Y", "fBodyAccJerk.mean...Z",
    "fBodyAccJerk.std...X", "fBodyAccJerk.std...Y", "fBodyAccJerk.std...Z")

这是我希望的结果:

 [3]"Time_Body_Acc_Mean_X"                "Time_Body_Acc_Mean_Y"               
 [5] "Time_Body_Acc_Mean_Z"                "Time_Body_Acc_Stddev_X"             
 [7] "Time_Body_Acc_Stddev_Y"              "Time_Body_Acc_Stddev_Z"             
 [9] "Time_Gravity_Acc_Mean_X"             "Time_Gravity_Acc_Mean_Y"            
[11] "Time_Gravity_Acc_Mean_Z"             "Time_Gravity_Acc_Stddev_X"          
[13] "Time_Gravity_Acc_Stddev_Y"           "Time_Gravity_Acc_Stddev_Z"

...

[43] "Freq_Body_Acc_Mean_X"                "Freq_Body_Acc_Mean_Y"               
[45] "Freq_Body_Acc_Mean_Z"                "Freq_Body_Acc_Stddev_X"             
[47] "Freq_Body_Acc_Stddev_Y"              "Freq_Body_Acc_Stddev_Z"             
[49] "Freq_Body_Acc_Jerk_Mean_X"           "Freq_Body_Acc_Jerk_Mean_Y"          
[51] "Freq_Body_Acc_Jerk_Mean_Z"           "Freq_Body_Acc_Jerk_Stddev_X"        
[53] "Freq_Body_Acc_Jerk_Stddev_Y"         "Freq_Body_Acc_Jerk_Stddev_Z"

我用正则表达式编写了一种非常冗长的更改方法。

vars_list <- unlist(lapply(vars_list, function(x){gsub("^t", "Time", x)}))
vars_list <- unlist(lapply(vars_list, function(x){gsub("^f", "Freq", x)}))
vars_list <- unlist(lapply(vars_list, function(x){gsub("std", "Stddev", x)}))
vars_list <- unlist(lapply(vars_list, function(x){gsub("mean", "Mean", x)}))
vars_list <- unlist(lapply(vars_list, function(x){gsub("\\.+", "", x)}))
vars_list <- unlist(lapply(vars_list, function(x){gsub("\\.", "", x)}))
vars_list <- unlist(lapply(vars_list, 
                           function(x){gsub("(?<=[a-z]).{0}(?=[A-Z])",
                                            "_", x, perl = TRUE)}))

有没有办法通过在单个函数调用中包含两个或多个格式化步骤来更高效、更优雅地获得相同的结果?

2 回复 | 直到 8 年前

akuiper 8 年前

另一种选择是写你的 patterns 和 replacement 在两个向量中,然后使用 stringi::stri_replace_all_regex 它可以以矢量化的方式进行替换:

# patterns correspond to replacement at the same positions
patterns <- c('^t', '^f', 'std', 'mean', '\\.+', '(?<=[a-z])([A-Z])')
replacement <- c('Time', 'Freq', 'Stddev', 'Mean', '', '_$1')

library(stringi)
stri_replace_all_regex(vars_list, patterns, replacement, vectorize_all = F)
# [1] "Time_Body_Acc_Mean_X"      "Time_Body_Acc_Mean_Y"     
# [3] "Time_Body_Acc_Mean_Z"      "Time_Body_Acc_Stddev_X"   
# [5] "Time_Body_Acc_Stddev_Y"    "Time_Body_Acc_Stddev_Z"   
# [7] "Time_Gravity_Acc_Mean_X"   "Time_Gravity_Acc_Mean_Y"  
# [9] "Time_Gravity_Acc_Mean_Z"   "Time_Gravity_Acc_Stddev_X"
#[11] "Time_Gravity_Acc_Stddev_Y" "Time_Gravity_Acc_Stddev_Z"

Maurits Evers 8 年前

这个用R基怎么样 sub ?

sub("t(\\w+)(Acc)\\.(\\w+)\\.+([XYZ])", "Time_\\1_\\2_\\3_\\4", vars_list);
#[1] "Time_Body_Acc_mean_X"    "Time_Body_Acc_mean_Y"
#[3] "Time_Body_Acc_mean_Z"    "Time_Body_Acc_std_X"
#[5] "Time_Body_Acc_std_Y"     "Time_Body_Acc_std_Z"
#[7] "Time_Gravity_Acc_mean_X" "Time_Gravity_Acc_mean_Y"
#[9] "Time_Gravity_Acc_mean_Z" "Time_Gravity_Acc_std_X"
#[11] "Time_Gravity_Acc_std_Y"  "Time_Gravity_Acc_std_Z"

改变 mean 到 Mean 和 std 到 StdDev 需要另外两个 附属的 s 同上 t 到 Time 和 f 到 Freq .