Question

我有一个包含多个列的df，其值前面是＆＃34; $＆＃34;像这样：

> str(data)
Classes ‘data.table’ and 'data.frame':  196879 obs. of  32 variables:
 $ City             : chr  "" "" "" "" ...
 $ Company_Goal     : chr  "" "" "" "" ...
 $ Company_Name     : chr  "" "" "" "" ...
 $ Event_Date       : chr  "5/14/2016" "9/26/2015" "9/12/2015" "6/3/2017" ...
 $ Event_Year       : chr  "FY 2016" "FY 2016" "FY 2016" "FY 2017" ...
 $ Fundraising_Goal : chr  "$250" "$200" "$350" "$0" ...
 $ Name             : chr  "Heart Walk 2015-2016 St. Louis MO" "Heart Walk 2015-2016 Canton, OH" "Heart Walk 2015-2016 Dallas, TX" "FDA HW 2016-2017 Albany, NY WO-65355" ...
 $ Participant_Id   : chr  "2323216" "2273391" "2419569" "4088558" ...
 $ State            : chr  "" "OH" "TX" "" ...
 $ Street           : chr  "" "" "" "" ...
 $ Team_Average     : chr  "$176" "$123" "$306" "$47" ...
 $ Team_Captain     : chr  "No" "No" "Yes" "No" ...
 $ Team_Count       : chr  "7" "6" "4" "46" ...
 $ Team_Id          : chr  "152788" "127127" "45273" "179207" ...
 $ Team_Member_Goal : chr  "$0" "$0" "$0" "$0" ...
 $ Team_Name        : chr  "Team Clayton" "Cardiac Crusaders" "BIS - Team Myers" "Independent Walkers" ...
 $ Team_Total_Gifts : chr  "$1,230 " "$738" "$1,225 " "$2,145 " ...
 $ Zip              : chr  "" "" "" "" ...
 $ Gifts_Count      : chr  "2" "1" "2" "1" ...
 $ Registration_Gift: chr  "No" "No" "No" "No" ...
 $ Participant_Gifts: chr  "$236" "$218" "$225" "$0" ...
 $ Personal_Gift    : chr  "$0" "$0" "$0" "$250" ...
 $ Total_Gifts      : chr  "$236" "$218" "$225" "$250" ...
 $ MATCH_CODE       : chr  "UX000" "UX000" "UX000" "UX000" ...
 $ TAP_LEVEL        : chr  "X" "X" "X" "X" ...
 $ TAP_DESC         : chr  "" "" "" "" ...
 $ TAP_LIFED        : chr  "" "" "" "" ...
 $ MEDAGE_CY        : chr  "0" "0" "0" "0" ...
 $ DIVINDX_CY       : chr  "0" "0" "0" "0" ...
 $ MEDHINC_CY       : chr  "0" "0" "0" "0" ...
 $ MEDDI_CY         : chr  "0" "0" "0" "0" ...
 $ MEDNW_CY         : chr  "0" "0" "0" "0" ...
 - attr(*, ".internal.selfref")=<externalptr>

我正在尝试删除所有＆＃34; $＆＃34;。我无法这样做 - 我已经尝试了this post以及this one中提供的建议，但在这两种情况下 - 数据保持不变......

帮助？

Answer 1

美元符号是正则表达式中的保留字符（有关详细信息，请参阅here）。 gsub()函数假定pattern默认为正则表达式。

您必须使用反斜杠（\\$）转义美元符号以匹配文字$。

#sample data
df = data.frame(Team_Average = c("$176", "$123", "$306"),
                Name = c("Heart Walk 2015-2016 St. Louis MO", 
                         "Heart Walk 2015-2016 Canton, OH",
                         "Heart Walk 2015-2016 Dallas, TX"),
                stringsAsFactors = FALSE)

df[] = lapply(df, gsub, pattern="\\$", replacement="")

或者，您可以使用gsub的{{1}}选项与字面上的fixed=TRUE匹配。

pattern

Answer 2

其他答案很好地适用于提供的示例。但是，如果数据集包含任何数字列，则通过stringr::str_replace_all()运行lapply()或library(stringr) library(dplyr) d <- data_frame( x = c("$200", "$191.40", "80.12"), y = c("$test", "column", "$foo"), z = 1:3 ) d[] <- lapply(d, gsub, pattern = "\\$", replacement = "") # A tibble: 3 x 3 x y z <chr> <chr> <chr> 1 200 test 1 2 191.40 column 2 3 80.12 foo 3会将数字列与字符串联：

请注意上面$的课程。

以下是从所有字符列中删除d %>% mutate_if( is.character, funs(str_replace_all(., "\\$", "")) ) # A tibble: 3 x 3 x y z <chr> <chr> <int> 1 200 test 1 2 191.40 column 2 3 80.12 foo 3的整合方法：

    public function __construct()
    {
            $this->middleware('auth:admin');
           // $this->middleware('auth:employer');
    }

删除所有＆＃34; $＆＃34;来自整个数据框架

2 个答案: