如何从 ls 中总结文件大小，如输出日志与字节、KiB、MiB、GiB

Question

我有一个预先计算的ls类似输出（它不是来自实际ls命令），我无法修改或重新计算它。它看起来像这样：

2016-10-14 14:52:09    0 Bytes folder/
2020-04-18 05:19:04  201 Bytes folder/file1.txt
2019-10-16 00:32:44  201 Bytes folder/file2.txt
2019-08-26 06:29:46  201 Bytes folder/file3.txt
2020-07-08 16:13:56  411 Bytes folder/file4.txt
2020-04-18 03:03:34  201 Bytes folder/file5.txt
2019-10-16 08:27:11    1.1 KiB folder/file6.txt
2019-10-16 10:13:52  201 Bytes folder/file7.txt
2019-10-16 08:44:35  920 Bytes folder/file8.txt
2019-02-17 14:43:10  590 Bytes folder/file9.txt

该日志可能有GiB，MiB，KiB，Bytes至少。其中可能的值是零值，或每个前缀的值 w/wo 逗号：

0   Bytes
3.9 KiB
201 Bytes
2.0 KiB
2.7 MiB
1.3 GiB

类似的方法如下

awk 'BEGIN{ pref[1]="K";  pref[2]="M"; pref[3]="G";} { total = total + $1; x = $1; y = 1; while( x  > 1024 ) { x = (x + 1023)/1024; y++; }  printf("%g%s\t%s\n",int(x*10)/10,pref[y],$2); } END { y = 1; while(  total > 1024 ) { total = (total + 1023)/1024; y++; } printf("Total:  %g%s\n",int(total*10)/10,pref[y]); }'

但在我的情况下不能正常工作：

$ head -n 10 files_sizes.log | awk '{print $3,$4}' | sort | awk 'BEGIN{ pref[1]="K";  pref[2]="M"; pref[3]="G";} { total = total + $1; x = $1; y = 1; while( x  > 1024 ) { x = (x + 1023)/1024; y++; }  printf("%g%s\t%s\n",int(x*10)/10,pref[y],$2); } END { y = 1; while(  total > 1024 ) { total = (total + 1023)/1024; y++; } printf("Total:  %g%s\n",int(total*10)/10,pref[y]); }'


0K  Bytes
1.1K    KiB
201K    Bytes
201K    Bytes
201K    Bytes
201K    Bytes
201K    Bytes
411K    Bytes
590K    Bytes
920K    Bytes
Total:  3.8M

此输出错误地计算了大小。我想要的输出是输入日志文件的正确总和：

0 Bytes
201 Bytes
201 Bytes
201 Bytes
411 Bytes
201 Bytes
1.1 KiB
201 Bytes
920 Bytes
590 Bytes
Total:  3.95742 KiB

笔记

求和的正确值Bytes是 201 * 5 + 590 + 920 = 2926，所以总和KiB是 2.857422 + 1.1 = 3,95742 KiB = 4052.400 Bytes

[更新]

我更新了KamilCuk和 Ted Lyngmo和Walter A解决方案的结果比较，这些解决方案给出了几乎相同的值：

$ head -n 10 files_sizes.log | tr -s ' ' | cut -d' ' -f3,4 | sed 's/ //; s/Bytes//; s/B//' | gnumfmt --from=auto | awk '{s+=$1}END{print s " Bytes"}'
117538 Bytes
$ head -n 1000 files_sizes.log | tr -s ' ' | cut -d' ' -f3,4 | sed 's/ //; s/Bytes//; s/B//' | gnumfmt --from=auto | awk '{s+=$1}END{print s " Bytes"}'
1225857 Bytes
$ head -n 10000 files_sizes.log | tr -s ' ' | cut -d' ' -f3,4 | sed 's/ //; s/Bytes//; s/B//' | gnumfmt --from=auto | awk '{s+=$1}END{print s " Bytes"}'
12087518 Bytes
$ head -n 1000000 files_sizes.log | tr -s ' ' | cut -d' ' -f3,4 | sed 's/ //; s/Bytes//; s/B//' | gnumfmt --from=auto | awk '{s+=$1}END{print s " Bytes"}'
77238840381 Bytes
$ head -n 100000000 files_sizes.log | tr -s ' ' | cut -d' ' -f3,4 | sed 's/ //; s/Bytes//; s/B//' | gnumfmt --from=auto | awk '{s+=$1}END{print s " Bytes"}'
2306569381835 Bytes

和

$ head -n 10 files_sizes.log | ./count_files.sh
3.957422 KiB
$ head -n 1000 files_sizes.log | ./count_files.sh
1.168946 MiB
$ head -n 10000 files_sizes.log | ./count_files.sh
11.526325 MiB
$ head -n 1000000 files_sizes.log | ./count_files.sh
71.934024 GiB
$ head -n 100000000 files_sizes.log | ./count_files.sh
2.097807 TiB

和

(head -n 100000000 files_sizes.log | tr -s ' ' | cut -d' ' -f3,4 | sed 's/Bytes//; s/KiB/* 1024/; s/MiB/* 1024 * 1024/;s/GiB/* 1024 * 1024 * 1024/; s/$/ + /; $s/+ //' | tr -d '\n' ; echo) | bc
2306563692898.8

在哪里

2.097807 TiB = 2.3065631893 TB = 2306569381835 字节

在计算上，我比较了所有三种解决方案的速度：

$ time head -n 100000000 files_sizes.log | ./count_files.sh
2.097807 TiB

real    2m7.956s
user    2m10.023s
sys 0m1.696s

$ time head -n 100000000 files_sizes.log | tr -s ' ' | cut -d' ' -f3,4 | sed 's/ //; s/Bytes//; s/B//' | gnumfmt --from=auto | awk '{s+=$1}END{print s " Bytes"}'
2306569381835 Bytes

real    4m12.896s
user    5m45.750s
sys 0m4.026s

$ time (head -n 100000000 files_sizes.log | tr -s ' ' | cut -d' ' -f3,4 | sed 's/Bytes//; s/KiB/* 1024/; s/MiB/* 1024 * 1024/;s/GiB/* 1024 * 1024 * 1024/; s/$/ + /; $s/+ //' | tr -d '\n' ; echo) | bc
2306563692898.8

real    4m31.249s
user    6m40.072s
sys 0m4.252s

Answer 1

使用numfmt这些数字转换。

cat <<EOF |
2016-10-14 14:52:09    0 Bytes folder/
2020-04-18 05:19:04  201 Bytes folder/file1.txt
2019-10-16 00:32:44  201 Bytes folder/file2.txt
2019-08-26 06:29:46  201 Bytes folder/file3.txt
2020-07-08 16:13:56  411 Bytes folder/file4.txt
2020-04-18 03:03:34  201 Bytes folder/file5.txt
2019-10-16 08:27:11    1.1 KiB folder/file6.txt
2019-10-16 10:13:52  201 Bytes folder/file7.txt
2019-10-16 08:44:35  920 Bytes folder/file8.txt
2019-02-17 14:43:10  590 Bytes folder/file9.txt
2019-02-17 14:43:10  3.9 KiB  folder/file9.txt
2019-02-17 14:43:10  2.7 MiB folder/file9.txt
2019-02-17 14:43:10  1.3 GiB folder/file9.txt
EOF
# extract 3rd and 4th column
tr -s ' ' | cut -d' ' -f3,4 |
# Remove space, remove "Bytes", remove "B"
sed 's/ //; s/Bytes//; s/B//' |
# convert to numbers
numfmt --from=auto |
# sum
awk '{s+=$1}END{print s}'

输出总和。

Answer 2

对于如下所述的输入：

2016-10-14 14:52:09    0 Bytes folder/
2020-04-18 05:19:04  201 Bytes folder/file1.txt
2019-10-16 00:32:44  201 Bytes folder/file2.txt
2019-08-26 06:29:46  201 Bytes folder/file3.txt
2020-07-08 16:13:56  411 Bytes folder/file4.txt
2020-04-18 03:03:34  201 Bytes folder/file5.txt
2019-10-16 08:27:11    1.1 KiB folder/file6.txt
2019-10-16 10:13:52  201 Bytes folder/file7.txt
2019-10-16 08:44:35  920 Bytes folder/file8.txt
2019-02-17 14:43:10  590 Bytes folder/file9.txt

您可以使用您希望能够解码的单位表：

BEGIN {
    unit["Bytes"] = 1;

    unit["kB"] = 10**3;
    unit["MB"] = 10**6;
    unit["GB"] = 10**9;
    unit["TB"] = 10**12;
    unit["PB"] = 10**15;
    unit["EB"] = 10**18;
    unit["ZB"] = 10**21;
    unit["YB"] = 10**24;

    unit["KB"] = 1024;
    unit["KiB"] = 1024**1;
    unit["MiB"] = 1024**2;
    unit["GiB"] = 1024**3;
    unit["TiB"] = 1024**4;
    unit["PiB"] = 1024**5;
    unit["EiB"] = 1024**6;
    unit["ZiB"] = 1024**7;
    unit["YiB"] = 1024**8;
}

然后在主循环中总结一下：

{
    if($4 in unit) total += $3 * unit[$4];
    else printf("ERROR: Can't decode unit at line %d: %s\n", NR, $0);
}

并在最后打印结果：

END {
    binaryunits[0] = "Bytes";
    binaryunits[1] = "KiB";
    binaryunits[2] = "MiB";
    binaryunits[3] = "GiB";
    binaryunits[4] = "TiB";
    binaryunits[5] = "PiB";
    binaryunits[6] = "EiB";
    binaryunits[7] = "ZiB";
    binaryunits[8] = "YiB";
    for(i = 8;; --i) {
         if(total >= 1024**i || i == 0) {
            printf("%.3f %s\n", total/(1024**i), binaryunits[i]);
            break;
        }
    }
}

输出：

3.957 KiB

请注意，您可以将 she-bang 添加到 awk 脚本的开头，以便可以单独运行它，这样您就不必将其放入 bash 脚本中：

#!/usr/bin/awk -f