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chess_freak
Asked: 2019-12-12 13:00:38 +0800 CST

如何操作特定字段？

  • 772

我有一个 csv 文件。它的行号最大为 25。

标题是时间（UTC）（第 1 场）、纬度（第 2 场）、经度（第 3 场）、深度（第 4 场）、mag（第 5 场）、地点（第 14 场）等

样本数据

2019-12-10T21:58:28.816Z 35.488 26.4157 57.32 5.4 35km NNE of Palaikastron, Greece


2019-12-11T11:54:27.670Z 18.6158 -67.2838 85 2.85 23km NW of San Antonio, Puerto Rico

首先，我要插入标题，包括内置变量 fieldwith。其次，我想将time(utc)转换为utc+03:00并分为标题的日期和时间并更改其日期格式。第三，我想在第 14 个字段的国家名称之前提取 ' of ' 单词和逗号之间的匹配项。

所需的输出为标题：

Date Time Latitude Longitude Depth Mag Place

期望的输出：

11.12.2019 00:58:28 35.488 26.4157 57.32 5.4 Palaikastron


11.12.2019 14:54:27 18.6158 -67.2838 85 2.85 San Antonio

time,latitude,longitude,depth,mag,magType,nst,gap,dmin,rms,net,id,updated,place,type,horizontalError,depthError,magError,magNst,status,locationSource,magSource
2019-12-06T13:04:46.931Z,-15.2838,-175.1193,10,6,mww,,50,3.512,0.81,us,us60006n19,2019-12-07T13:11:48.228Z,"164km WNW of Hihifo, Tonga",earthquake,8.4,1.9,0.08,15,reviewed,us,us
2019-12-04T20:10:03.614Z,-19.0515,169.5628,266,6,mww,,21,2.812,0.82,us,us60006m2j,2019-12-05T23:44:01.300Z,"63km NNE of Isangel, Vanuatu",earthquake,7.6,1.9,0.037,71,reviewed,us,us
2019-12-03T08:46:36.374Z,-18.5597,-70.6504,32.44,6,mww,,112,0.31,1.4,us,us70006fh7,2019-12-05T08:07:29.617Z,"37km WSW of Arica, Chile",earthquake,6.2,7.8,0.069,20,reviewed,us,us
2019-12-02T05:01:54.693Z,51.3218,-178.2425,27.33,6,mww,,104,0.862,0.97,us,us70006f6d,2019-12-07T02:09:55.119Z,"60km E of Amatignak Island, Alaska",earthquake,6.7,4.2,0.066,22,reviewed,us,us
2019-11-27T07:23:42.552Z,35.7272,23.2673,71.76,6,mww,,23,1.394,1.16,us,us70006dlt,2019-12-03T23:18:27.456Z,"41km NW of Platanos, Greece",earthquake,5.8,5.4,0.046,46,reviewed,us,us
2019-11-26T02:54:12.594Z,41.5112,19.5151,20,6.4,mww,,17,0.937,0.58,us,us70006d0m,2019-12-09T15:46:11.689Z,"16km WSW of Mamurras, Albania",earthquake,3.5,1.8,0.037,72,reviewed,us,us
2019-11-24T00:54:01.052Z,51.3809,-175.5108,20,6.3,mww,,22,0.658,0.95,us,us70006cb6,2019-12-10T01:04:03.731Z,"96km SE of Adak, Alaska",earthquake,3.9,1.8,0.05,38,reviewed,us,us
2019-11-23T12:11:16.261Z,1.6286,132.7854,10,6.1,mww,,38,4.549,1.1,us,us70006c6w,2019-11-25T21:00:33.040Z,"Papua region, Indonesia",earthquake,7.8,1.8,0.061,26,reviewed,us,us
2019-11-20T23:50:43.955Z,19.4533,101.3558,10,6.2,mww,,15,2.366,0.62,us,us70006ara,2019-12-04T05:52:37.313Z,"32km ESE of Chaloem Phra Kiat, Thailand",earthquake,6.4,1.7,0.049,40,reviewed,us,us

如果我能成功一件事，我就不会做其他事。这对我来说是挑战。 请指导我。一方面，难以理解awk为它使用。另一方面，awk时间函数非常有用。我现在很困惑。无论我尝试过什么，我都不擅长。

command-line bash date awk

2 个回答

  1. 首先，您需要预处理 CSV 输入以更好地处理嵌入的逗号。然后将 AWK 分解为功能块。

    $ cat preprocess.sed
#!/bin/sed -f
:start   # loop back to here
/"/{  # for any line that has a double quote
  h   # copy to the hold buffer
  s/[^"]*"\([^"]*\).*/\1/  # what is between the first pair of dquotes
  s/,/@@/g    # replace comma with '@@'
  G   # append the hold buffer to the pattern buffer
      # so we get what was in dqoutes followed by a newline followed by the
      # original line
  s/\(.*\)\n\([^"]*\)"\([^"]*\)"\(.*\)/\2\1\4/
      # replace the unquoted part with what was there
  t start   # go back to 'start'
}

    这将替换".*,.*"with .*@@.*，这将使 AWK 更容易。

    要将日期更改为新时区，请替换第一行：

    $ cat change_date.sh
#!/bin/sh
userTZ="${1:-UTC+3}"
sed 's/,/ /' |
    while read datestr rest; do
        if [ "${datestr}" = time ]; then
            newdate="${datestr}"
        else
            newdate=$(TZ=${userTZ} date -d "${datestr}" "+%d %m %Y %H:%M:%S")
        fi
        echo "${newdate}:${rest}"

    done

    AWK 脚本如下所示：

    $ cat reformat.awk
#!/bin/awk  -f
BEGIN {IFS=","}  # comma separated fields
NR==1 {print; next;}  # print the header and do nothing more with it
{   # get just the "town" from the place field
    sub(/.* of /,"",$14)  # strip up to the " of "
    sub(/@@ .*/,"",$14)   # strip after the embedded comma (now '@@')
}
{
    printf("%s %8.3f %8.3f %8.3fs %8.3f %s\n", $1, $2, $3, $4, $5, $14)
}

    确保两者都是可执行并运行preprocess.sed sample.csv | change_date.sh | reformat.awk

    或者在一行：

    sed ':start;/"/{;h;s/[^"]*"\([^"]*\).*/\1/;s/,/@@/g;G;s/\(.*\)\n\([^"]*\)"\([^"]*\)"\(.*\)/\2\1\4/;t start;};s/,/ /' test.csv | while read datestr rest; do if [ "$datestr" = "time" ]; then newdate="${datestr}"; else newdate=$(TZ=UTC+3 date -d "$datestr" "+%d %m %Y %H:%M:%S"); fi; echo "${newdate},${rest}"; done | awk -F, 'NR==1 {print;next} {sub(/.* of /,"",$14);sub(/@@ .*/,"",$14)} {printf("%s %8.3f %8.3f %8.3fs %8.3f %s\n", $1, $2, $3, $4, $5, $14)}'
    • 2
  2. Best Answer

    尽管学习 awk 是一个令人钦佩的目标，但它没有用于解析真正的 CSV 文件（特别是可能包含转义或引用分隔符的字段）的内置机制 - 并且时间函数是 GNU 特定的且不可移植。

    由于这些原因，您可能需要考虑使用 Perl（及其Text::CSV模块）、Python - 或者我目前最喜欢的这类东西，Miller。除了提供真正的 CSV 解析外，它们还提供了适当的strptime功能，而即使使用 GNU awk，mktime您也需要手动解析和组合datespec参数。

    例如，在 Miller 中，您可以执行以下操作：

    mlr --csv \
  put -S '
    s = strptime($time,"%Y-%m-%dT%H:%M:%SZ") + 3*3600; 
    $date = strftime(s,"%d.%m.%Y"); 
    $time = strftime(s,"%H:%M:%S"); 
    $place =~ "(.* of |)([^,]*),(.*)$" { $place = "\2" }
  ' then cut -o -f date,time,latitude,longitude,depth,mag,place input.csv

    如果您想要空格分隔的输出列，请更改--csv为--icsv --opprint（“漂亮打印的”表格输出 - 带有标题）或--icsv --onidx（简单的空格分隔输出）。

    前任。

    $ mlr --icsv --opprint   put -S '
    s = strptime($time,"%Y-%m-%dT%H:%M:%SZ") + 3*3600; 
    $date = strftime(s,"%d.%m.%Y"); 
    $time = strftime(s,"%H:%M:%S"); 
    $place =~ "(.* of |)([^,]*),(.*)$" { $place = "\2" }
  ' then cut -o -f date,time,latitude,longitude,depth,mag,place input.csv
date       time     latitude longitude depth mag place
06.12.2019 16:04:46 -15.2838 -175.1193 10    6   Hihifo
04.12.2019 23:10:03 -19.0515 169.5628  266   6   Isangel
03.12.2019 11:46:36 -18.5597 -70.6504  32.44 6   Arica
02.12.2019 08:01:54 51.3218  -178.2425 27.33 6   Amatignak Island
27.11.2019 10:23:42 35.7272  23.2673   71.76 6   Platanos
26.11.2019 05:54:12 41.5112  19.5151   20    6.4 Mamurras
24.11.2019 03:54:01 51.3809  -175.5108 20    6.3 Adak
23.11.2019 15:11:16 1.6286   132.7854  10    6.1 Papua region
21.11.2019 02:50:43 19.4533  101.3558  10    6.2 Chaloem Phra Kiat

    Miller 可从 Ubuntuuniverse存储库获得。

    • 2

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