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主页 / unix / 问题

问题[data](unix)

Martin Hope
Lucas Djeu
Asked: 2022-01-28 18:57:28 +0800 CST

从多个文件中平均第 n 行到一个平均主文件

  • -1

我有 3 个文件,其中包含 8 行数值和文本。我试图取所有三个文件中每一行的平均值,并用这些平均值打印一个新文件。下面是三个示例文件,都是同名格式 testfile1.1, testfile1.2, testfile1.3

测试文件1.1

1
2048
8
5
5
4
9
Lat:1

测试文件1.2

1
2048
10
7
7
4
9
Lat:1

测试文件1.3

1
2048
3
6
3
4
6
Lat:7

我希望输出文件如下所示(取平均值后)

平均文件1

1
2048
7
6
5
4
8
Lat:3

希望这对我正在尝试做的事情有意义!

我尝试过使用 awk、sed 的不同组合,它们对 3-4 行数据运行良好,但我的实际数据在 40 多个文件名中有 2000 多行

编辑:所以我能够理解如何控制我想要打印的 sig figs 以及如何编辑正则表达式以更好地匹配浮动小数。

(请让我知道我是否应该将此作为另一个问题并删除此问题!)。

我的实际数据有很多其他行,其中包含文本以及我想要取平均值的值。我试图创建额外的字符串,但后来变得更加困惑。在我的真实文件中,在某些行上,我想要不同的命令,例如从行中打印文本,对实际数据取平均值,复制文本的行和数据的平均值以及日期和时间的平均值。

下面是 2 个文件的副本(每行都有我想对它们做的评论)。

豆腐1.1

ABCDEFGH #print text into output file (same on both files)
1     # Take average of values across all the files in this line
2048  # Take average of values across all the files in this line
8     # Take average of values across all the files in this line
5     # Take average of values across all the files in this line
5     # Take average of values across all the files in this line
4     # Take average of values across all the files in this line
9.5   # Take average of values across all the files in this line
1     # Take average of values across all the files in this line
90.00  # Check and make sure value in this line across print if same
Sprite # check and see if text is same across all values and print if same
cats10   # check and see if text is same across all values and print if same
07/02/20 # See below for explantion on next 3 lines
08:32
08:32
290.000000 # average across all 3 files on this line
10.750000 # average across all 3 files on this line
SCANS23   # output should be SCANS "average of values"
INT_TIME57500 # output should be INT_TIME with sum of all values
SITE northpole   #Check if all lines are same if so print line
LONGITUDE -147.850037  # Output should be LONGITUDE%f
LATITUDE 64.859375     # Output should be LONGITUDE%f

第 13 行是数据的来源日期,第 14 行是开始时间和结束时间。可能使用某种日期到十进制命令..有没有办法取日期的平均值?如果一个数据是在 2020 年 7 月 2 日获取的,而另一个数据是在 2018 年 7 月 2 日获取的,那么输出可以是 19 年 7 月 2 日吗?时间的平均值也会被考虑在内。

豆腐1.2

ABCDEFGH #print text into output file (same on both files)
1     # Take average of values across all the files in this line
2048  # Take average of values across all the files in this line
10    # Take average of values across all the files in this line
7     # Take average of values across all the files in this line
7     # Take average of values across all the files in this line
4     # Take average of values across all the files in this line
8   # Take average of values across all the files in this line
1     # Take average of values across all the files in this line
90.00  # Check and make sure value in this line across print if same
Sprite # check and see if text is same across all values and print if same
cats10   # check and see if text is same across all values and print if same
07/02/20 # See below for explanation on next 3 lines
08:32
08:32
290.000000 # average across all 3 files on this line
10.750000 # average across all 3 files on this line
SCANS23   # output should be SCANS "average of values"
INT_TIME57500 # output should be INT_TIME with sum of all values
SITE northpole   #Check if all lines are same if so print line
LONGITUDE -147.850037  # Output should be LONGITUDE%f
LATITUDE 64.859375     # Output should be LONGITUDE%f

我厌倦了尝试在脚本中包含多个字符串起始值的尝试,但很快就变得非常混乱。

awk -F: '
  FNR==1     { c++ };
  /^LATITUDE/    { a[FNR] += $6 };
  /^SCANS/    { a[FNR] += $2 };
  /^[+-]?([0-9]*[.])?[0-9]+$/ { a[FNR] += $1 };

  END {
    for (i in a) {
      printf (i==22 ? "LATITUDE%f": i==18 ? "SCANS%2.3f": "%f") "\n", a[i] / c
    }
  }' tofu1.* > askforhelp

这给了我

$ more askforhelp

90.000000
LATITUDE0.000000
290.000000
10.750000
SCANS0.000
1.000000
2048.000000
6.333333
4.666667
5.000000
4.000000
7.833333
2.666667

我还尝试一次添加多个文本字符串,当我完全没有从这次尝试中得到输出时,我感到非常困惑。

awk -F: '
  FNR==1     { c++ };
  /^LATITUDE/    { a[FNR] += $6 };
  /^LONGITUDE/    { a[FNR] += $5 };
  /^SITE/    { a[FNR] += $4 };
  /^INT_TIME/    { a[FNR] += $3 };
  /^SCANS/    { a[FNR] += $2 };
  /^[+-]?([0-9]*[.])?[0-9]+$/ { a[FNR] += $1 };

  END {
    for (i in a) {
      printf (i==22 ? "LATITUDE%f": 
              i==21 ? "LONGITUDE%2.3f": 
              i==20 ? "SITE%2.3f": 
              i==19 ? "INT_TIME%2.3f": 
              i==18 ? "SCANS%2.3f": "%f") "\n", a[i] / c 
    }
  }' /home/lmdjeu/test/test1.* > /home/lmdjeu/test/askforhelp

data math
  • 1 个回答
  • 72 Views
Martin Hope
cantik memekmu
Asked: 2021-10-31 16:28:10 +0800 CST

复制整个目录结构和/或骨架

  • 0

我们如何复制整个目录模板/结构或骨架?
由于没有使用说明:

Copy SOURCE to DEST, or multiple SOURCE(s) to DIRECTORY.

  -a, --archive                same as -dR --preserve=all
      --attributes-only        don't copy the file data, just the attributes
      --backup[=CONTROL]       make a backup of each existing destination file
  -b                           like --backup but does not accept an argument
      --copy-contents          copy contents of special files when recursive
  -d                           same as --no-dereference --preserve=links
  -f, --force                  if an existing destination file cannot be
                                 opened, remove it and try again (this option
                                 is ignored when the -n option is also used)
  -i, --interactive            prompt before overwrite (overrides a previous -n
                                  option)
  -H                           follow command-line symbolic links in SOURCE
  -l, --link                   hard link files instead of copying
  -L, --dereference            always follow symbolic links in SOURCE
  -n, --no-clobber             do not overwrite an existing file (overrides
                                 a previous -i option)
  -P, --no-dereference         never follow symbolic links in SOURCE
  -p                           same as --preserve=mode,ownership,timestamps
      --preserve[=ATTR_LIST]   preserve the specified attributes (default:
                                 mode,ownership,timestamps), if possible
                                 additional attributes: context, links, xattr,
                                 all
      --no-preserve=ATTR_LIST  don't preserve the specified attributes
      --parents                use full source file name under DIRECTORY
  -R, -r, --recursive          copy directories recursively
      --reflink[=WHEN]         control clone/CoW copies. See below
      --remove-destination     remove each existing destination file before
                                 attempting to open it (contrast with --force)
      --sparse=WHEN            control creation of sparse files. See below
      --strip-trailing-slashes  remove any trailing slashes from each SOURCE
                                 argument
  -s, --symbolic-link          make symbolic links instead of copying
  -S, --suffix=SUFFIX          override the usual backup suffix
  -t, --target-directory=DIRECTORY  copy all SOURCE arguments into DIRECTORY
  -T, --no-target-directory    treat DEST as a normal file
  -u, --update                 copy only when the SOURCE file is newer
                                 than the destination file or when the
                                 destination file is missing
  -v, --verbose                explain what is being done
  -x, --one-file-system        stay on this file system
  -Z                           set SELinux security context of destination
                                 file to default type
      --context[=CTX]       like -Z, or if CTX is specified then set the SELinux 
                            or SMACK security context to CTX

请给出好的指导

filesystems data
  • 1 个回答
  • 263 Views
Martin Hope
PHB
Asked: 2019-11-23 06:43:33 +0800 CST

数据清理问题 - 用于删除尾随空格和逗号组合的正则表达式

  • 0

我有地址字符串(由于初始编程不佳)允许输入过多的自由文本。在清理这个继承的烂摊子的过程中,我注意到在地址字段的末尾,经常有(而且经常不是)我想去掉的逗号。

问题是(再次,经常!)那些输入数据的人(由线路支付......)输入的字符串如下:

`address_1_string  ,  `

那是 address_1_string 后跟两个空格,然后是我要删除的逗号,然后是另外两个空格。address_string_1空格的数量(在逗号之前或之后)是任意的,通常从 0 到 5。由于本身可能具有内部(有效)逗号这一事实,问题变得更加复杂。

所以,我正在寻找的是一个正则表达式,它会走到行尾,删除任何尾随空格、第一个逗号,然后是更多空格,直到有效的 [a-zA-Z0-9] ([:alphanum: ]?) 字符被找到。

我有正则表达式的概念,但这超出了我的工资等级。这是我在 Stackexchange 上的第一篇文章,所以如果我在错误的地方,请重定向我。TIA。

regular-expression data
  • 1 个回答
  • 140 Views
Martin Hope
topencrypt
Asked: 2019-09-05 03:10:02 +0800 CST

将磁盘克隆到更大的磁盘而不覆盖

  • -2

如果我将硬盘克隆到另一个(更大的)硬盘上已经有数据,旧数据会被删除还是会在旧数据之上添加新数据?

我需要将数据从 2 个较小的硬盘驱动器移动到 1 个较大的驱动器。将第一个驱动器克隆到更大的驱动器后,我想将第二个驱动器克隆到具有来自第一个驱动器的数据的同一个更大的驱动器。

来自第二个驱动器的数据(进入较大的硬盘)会覆盖来自第一个克隆的较大目标驱动器上的预先存在的数据,还是只是在较大的硬盘上预先存在的数据之上添加数据?

data external-hdd
  • 1 个回答
  • 754 Views
Martin Hope
user3776738
Asked: 2019-06-18 10:38:25 +0800 CST

如何删除文本文件中的重复行并获取已删除行的数量?

  • 0

我知道

awk '!seen[$0]++' filename > output.txt

可以从文本文件中删除所有重复项。但是我如何获取已删除行的列表,以便使用此信息删除另一个文件中的相同行。

我需要这个,因为我想删除用一种语言编写的大型文本文件中的所有重复项,然后使用该语言的翻译删除文件中的相同行,而不会丢失翻译的匹配。

为了更清楚:(文件 1)

line A
line B
line A
line C
...

(文件 2):

line 1
line 2
line 3
line 4
...

从文件 1 中删除“line A”并从文件 2 中删除“line 3”。需要翻译:

line A -> line 1
line B -> line 2
line C -> line 4
... 

文件 1B(已删除重复项)将如下所示:

line A
line B
line C
...

文件 2B(已删除重复项)将如下所示:

line 1
line 2
line 4
... 
text-processing data
  • 2 个回答
  • 486 Views
Martin Hope
neverMind9
Asked: 2019-01-20 11:15:58 +0800 CST

复制数千个文件后检查数据完整性[重复]

  • 15
这个问题在这里已经有了答案:
从一个硬盘驱动器复制到另一个硬盘驱动器后验证大目录 3 个答案
3年前关闭。

我将数千个文件复制到 exFAT MicroSD 卡中。

文件数和字节数相同,但我怎么知道数据是否损坏?

如果 JackPal Android 终端也支持该命令,那就太好了。

storage data
  • 5 个回答
  • 11202 Views
Martin Hope
neverMind9
Asked: 2018-10-12 18:08:40 +0800 CST

区分坏的逻辑块和物理块?(单独列出)

  • 0

工具badblocks可以提供不可读的 LBA 列表,包括我猜的逻辑错误。

如何区分逻辑(软)坏块和物理(硬)坏块?

  • 单独列出逻辑和物理错误或标记为。
  • 指示任何给定 LBA 的错误类型。
hard-disk data
  • 1 个回答
  • 510 Views
Martin Hope
Jackson Hart
Asked: 2018-07-29 08:59:07 +0800 CST

将文件合并为一个[重复]

  • 1
这个问题在这里已经有了答案:
并排打印两列中的两个文件 4 个答案
4年前关闭。

如果我有一个具有以下文件名的文件夹:

cluster_sizes_0.txt
cluster_sizes_1.txt
cluster_size_2.txt

等等

每个文件都包含一列值。

linux中是否有一个命令可以将所有文件合并到cluster_all.txt中?

这个新文件中的第一列将对应于cluster_sizes_0.txt,第二列将是cluster_sizes_1.txt等等。

可能有多达 200 个集群 txt 文件,但每个文件夹都会发生变化。我正在寻找一种方法来组合这些文件,而不是一个一个地复制。

另外,我需要确保它们按顺序粘贴到文件中。这可能与编号系统有一些问题,因为如果低于 10,我只包括个位数。

例如:

paste cluster_size.* > cluster_all.txt由于编号,不会按顺序粘贴它们。如何在不手动更改所有编号的情况下修复编号?

command-line data
  • 1 个回答
  • 72 Views
Martin Hope
stdcerr
Asked: 2018-07-13 14:33:36 +0800 CST

从 JSON 字符串中提取数据

  • 2

我必须提取一个车牌数字,它与 JSON 字符串相关的置信度如下所示:

{
  "response": {
    "container": {
      "id": "0df307bc-06b2-45cf-b7ff-ce07fd04e04d",
      "timestamp": "2018-Jul-10 17:34:27.448632"
    },
    "id": "00000002-0000-0000-0000-000000000015"
  },
  "frames": {
    "frame": {
      "id": "5583",
      "timestamp": "2016-Nov-30 13:05:27",
      "lps": {
        "lp": {
          "licenseplate": "15451BBL",
          "text": "15451BBL",
          "wtext": "15451BBL",
          "confidence": "20",
          "bkcolor": "16777215",
          "color": "16777215",
          "type": "0",
          "ntip": "11",
          "cct_country_short": "",
          "cct_state_short": "",
          "tips": {
            "tip": {
              "poly": {
                "p": { "x": "1094", "y": "643" },
                "p": { "x": "1099", "y": "643" },
                "p": { "x": "1099", "y": "667" },
                "p": { "x": "1094", "y": "667" }
              },
              "bkcolor": "16777215",
              "color": "0",
              "code": "49",
              "code_ascii": "1",
              "confidence": "97"
            },
            "tip": {
              "poly": {
                "p": { "x": "1103", "y": "642" },
                "p": { "x": "1113", "y": "642" },
                "p": { "x": "1112", "y": "667" },
                "p": { "x": "1102", "y": "667" }
              },
              "bkcolor": "16777215",
              "color": "0",
              "code": "53",
              "code_ascii": "5",
              "confidence": "89"
            },
            "tip": {
              "poly": {
                "p": { "x": "1112", "y": "640" },
                "p": { "x": "1122", "y": "640" },
                "p": { "x": "1122", "y": "666" },
                "p": { "x": "1112", "y": "666" }
              },
              "bkcolor": "16777215",
              "color": "0",
              "code": "52",
              "code_ascii": "4",
              "confidence": "97"
            },
            "tip": {
              "poly": {
                "p": { "x": "1123", "y": "640" },
                "p": { "x": "1132", "y": "640" },
                "p": { "x": "1131", "y": "665" },
                "p": { "x": "1123", "y": "665" }
              },
              "bkcolor": "16777215",
              "color": "0",
              "code": "53",
              "code_ascii": "5",
              "confidence": "97"
            },
            "tip": {
              "poly": {
                "p": { "x": "1134", "y": "640" },
                "p": { "x": "1139", "y": "640" },
                "p": { "x": "1139", "y": "664" },
                "p": { "x": "1133", "y": "664" }
              },
              "bkcolor": "16777215",
              "color": "0",
              "code": "49",
              "code_ascii": "1",
              "confidence": "77"
            },
            "tip": {
              "poly": {
                "p": { "x": "1154", "y": "639" },
                "p": { "x": "1163", "y": "639" },
                "p": { "x": "1163", "y": "663" },
                "p": { "x": "1153", "y": "663" }
              },
              "bkcolor": "16777215",
              "color": "0",
              "code": "66",
              "code_ascii": "B",
              "confidence": "97"
            },
            "tip": {
              "poly": {
                "p": { "x": "1164", "y": "638" },
                "p": { "x": "1173", "y": "638" },
                "p": { "x": "1173", "y": "663" },
                "p": { "x": "1163", "y": "663" }
              },
              "bkcolor": "16777215",
              "color": "0",
              "code": "66",
              "code_ascii": "B",
              "confidence": "94"
            },
            "tip": {
              "poly": {
                "p": { "x": "1191", "y": "637" },
                "p": { "x": "1206", "y": "636" },
                "p": { "x": "1205", "y": "660" },
                "p": { "x": "1190", "y": "661" }
              },
              "bkcolor": "16777215",
              "color": "0",
              "code": "76",
              "code_ascii": "L",
              "confidence": "34"
            },
            "tip": {
              "poly": {
                "p": { "x": "1103", "y": "655" },
                "p": { "x": "1111", "y": "655" },
                "p": { "x": "1111", "y": "667" },
                "p": { "x": "1103", "y": "667" }
              },
              "bkcolor": "16777215",
              "color": "0",
              "code": "74",
              "code_ascii": "J",
              "confidence": "57"
            },
            "tip": {
              "poly": {
                "p": { "x": "1103", "y": "655" },
                "p": { "x": "1111", "y": "655" },
                "p": { "x": "1111", "y": "667" },
                "p": { "x": "1103", "y": "667" }
              },
              "bkcolor": "16777215",
              "color": "0",
              "code": "74",
              "code_ascii": "J",
              "confidence": "57"
            },
            "tip": {
              "poly": {
                "p": { "x": "1176", "y": "638" },
                "p": { "x": "1185", "y": "637" },
                "p": { "x": "1184", "y": "661" },
                "p": { "x": "1175", "y": "662" }
              },
              "bkcolor": "16777215",
              "color": "0",
              "code": "52",
              "code_ascii": "4",
              "confidence": "7"
            }
          },
          "ncharacter": "8",
          "characters": {
            "characater": {
              "poly": {
                "p": { "x": "1094", "y": "643" },
                "p": { "x": "1099", "y": "643" },
                "p": { "x": "1099", "y": "667" },
                "p": { "x": "1094", "y": "667" }
              },
              "bkcolor": "16777215",
              "color": "0",
              "code": "49",
              "code_ascii": "1",
              "confidence": "97"
            },
            "characater": {
              "poly": {
                "p": { "x": "1103", "y": "642" },
                "p": { "x": "1113", "y": "642" },
                "p": { "x": "1112", "y": "667" },
                "p": { "x": "1102", "y": "667" }
              },
              "bkcolor": "16777215",
              "color": "0",
              "code": "53",
              "code_ascii": "5",
              "confidence": "89"
            },
            "characater": {
              "poly": {
                "p": { "x": "1112", "y": "640" },
                "p": { "x": "1122", "y": "640" },
                "p": { "x": "1122", "y": "666" },
                "p": { "x": "1112", "y": "666" }
              },
              "bkcolor": "16777215",
              "color": "0",
              "code": "52",
              "code_ascii": "4",
              "confidence": "97"
            },
            "characater": {
              "poly": {
                "p": { "x": "1123", "y": "640" },
                "p": { "x": "1132", "y": "640" },
                "p": { "x": "1131", "y": "665" },
                "p": { "x": "1123", "y": "665" }
              },
              "bkcolor": "16777215",
              "color": "0",
              "code": "53",
              "code_ascii": "5",
              "confidence": "97"
            },
            "characater": {
              "poly": {
                "p": { "x": "1134", "y": "640" },
                "p": { "x": "1139", "y": "640" },
                "p": { "x": "1139", "y": "664" },
                "p": { "x": "1133", "y": "664" }
              },
              "bkcolor": "16777215",
              "color": "0",
              "code": "49",
              "code_ascii": "1",
              "confidence": "77"
            },
            "characater": {
              "poly": {
                "p": { "x": "1154", "y": "639" },
                "p": { "x": "1163", "y": "639" },
                "p": { "x": "1163", "y": "663" },
                "p": { "x": "1153", "y": "663" }
              },
              "bkcolor": "16777215",
              "color": "0",
              "code": "66",
              "code_ascii": "B",
              "confidence": "97"
            },
            "characater": {
              "poly": {
                "p": { "x": "1164", "y": "638" },
                "p": { "x": "1173", "y": "638" },
                "p": { "x": "1173", "y": "663" },
                "p": { "x": "1163", "y": "663" }
              },
              "bkcolor": "16777215",
              "color": "0",
              "code": "66",
              "code_ascii": "B",
              "confidence": "94"
            },
            "characater": {
              "poly": {
                "p": { "x": "1191", "y": "637" },
                "p": { "x": "1206", "y": "636" },
                "p": { "x": "1205", "y": "660" },
                "p": { "x": "1190", "y": "661" }
              },
              "bkcolor": "16777215",
              "color": "0",
              "code": "76",
              "code_ascii": "L",
              "confidence": "34"
            }
          },
          "det_time_us": "1104009",
          "poly": {
            "p": { "x": "1088", "y": "642" },
            "p": { "x": "1210", "y": "634" },
            "p": { "x": "1210", "y": "661" },
            "p": { "x": "1087", "y": "669" }
          }
        }
      },
      "det_time_us": "1710270"
    }
  }
}

我有类似的东西:

$ jq -r '.frames.frame.lps.lp|.characters.characater.code_ascii,.characters.characater.confidence' test.json

但它只返回一个字母 & 它是 conf. 分数....

问题

  • 如何取回所有字母和相关分数?

输出

我希望输出如下:

1 97, 5 89, 4 97,5 97, 1 77,B 97, B 94, L 34, J, 57,J 57, 4 7, 1 97, 5 89, 4 97, 5 97, 1 77, B 97, B 94, L 34

注意:格式可以不同,这只是表示我要提取的数据。

复制粘贴

输入文件“test.json”

{"response":{"container":{"id":"41d6efcb-24d6-490d-8880-762255519b5f","timestamp":"2018-Jul-11 19:51:06.461665"},"id":"00000002-0000-0000-0000-000000000015"},"frames":{"frame":{"id":"5583","timestamp":"2016-Nov-30 13:05:27","lps":{"lp":{"licenseplate":"15451BBL","text":"15451BBL","wtext":"15451BBL","confidence":"20","bkcolor":"16777215","color":"16777215","type":"0","ntip":"11","cct_country_short":"","cct_state_short":"","tips":{"tip":{"poly":{"p":{"x":"1094","y":"643"},"p":{"x":"1099","y":"643"},"p":{"x":"1099","y":"667"},"p":{"x":"1094","y":"667"}},"bkcolor":"16777215","color":"0","code":"49","code_ascii":"1","confidence":"97"},"tip":{"poly":{"p":{"x":"1103","y":"642"},"p":{"x":"1113","y":"642"},"p":{"x":"1112","y":"667"},"p":{"x":"1102","y":"667"}},"bkcolor":"16777215","color":"0","code":"53","code_ascii":"5","confidence":"89"},"tip":{"poly":{"p":{"x":"1112","y":"640"},"p":{"x":"1122","y":"640"},"p":{"x":"1122","y":"666"},"p":{"x":"1112","y":"666"}},"bkcolor":"16777215","color":"0","code":"52","code_ascii":"4","confidence":"97"},"tip":{"poly":{"p":{"x":"1123","y":"640"},"p":{"x":"1132","y":"640"},"p":{"x":"1131","y":"665"},"p":{"x":"1123","y":"665"}},"bkcolor":"16777215","color":"0","code":"53","code_ascii":"5","confidence":"97"},"tip":{"poly":{"p":{"x":"1134","y":"640"},"p":{"x":"1139","y":"640"},"p":{"x":"1139","y":"664"},"p":{"x":"1133","y":"664"}},"bkcolor":"16777215","color":"0","code":"49","code_ascii":"1","confidence":"77"},"tip":{"poly":{"p":{"x":"1154","y":"639"},"p":{"x":"1163","y":"639"},"p":{"x":"1163","y":"663"},"p":{"x":"1153","y":"663"}},"bkcolor":"16777215","color":"0","code":"66","code_ascii":"B","confidence":"97"},"tip":{"poly":{"p":{"x":"1164","y":"638"},"p":{"x":"1173","y":"638"},"p":{"x":"1173","y":"663"},"p":{"x":"1163","y":"663"}},"bkcolor":"16777215","color":"0","code":"66","code_ascii":"B","confidence":"94"},"tip":{"poly":{"p":{"x":"1191","y":"637"},"p":{"x":"1206","y":"636"},"p":{"x":"1205","y":"660"},"p":{"x":"1190","y":"661"}},"bkcolor":"16777215","color":"0","code":"76","code_ascii":"L","confidence":"34"},"tip":{"poly":{"p":{"x":"1103","y":"655"},"p":{"x":"1111","y":"655"},"p":{"x":"1111","y":"667"},"p":{"x":"1103","y":"667"}},"bkcolor":"16777215","color":"0","code":"74","code_ascii":"J","confidence":"57"},"tip":{"poly":{"p":{"x":"1103","y":"655"},"p":{"x":"1111","y":"655"},"p":{"x":"1111","y":"667"},"p":{"x":"1103","y":"667"}},"bkcolor":"16777215","color":"0","code":"74","code_ascii":"J","confidence":"57"},"tip":{"poly":{"p":{"x":"1176","y":"638"},"p":{"x":"1185","y":"637"},"p":{"x":"1184","y":"661"},"p":{"x":"1175","y":"662"}},"bkcolor":"16777215","color":"0","code":"52","code_ascii":"4","confidence":"7"}},"ncharacter":"8","characters":{"characater":{"poly":{"p":{"x":"1094","y":"643"},"p":{"x":"1099","y":"643"},"p":{"x":"1099","y":"667"},"p":{"x":"1094","y":"667"}},"bkcolor":"16777215","color":"0","code":"49","code_ascii":"1","confidence":"97"},"characater":{"poly":{"p":{"x":"1103","y":"642"},"p":{"x":"1113","y":"642"},"p":{"x":"1112","y":"667"},"p":{"x":"1102","y":"667"}},"bkcolor":"16777215","color":"0","code":"53","code_ascii":"5","confidence":"89"},"characater":{"poly":{"p":{"x":"1112","y":"640"},"p":{"x":"1122","y":"640"},"p":{"x":"1122","y":"666"},"p":{"x":"1112","y":"666"}},"bkcolor":"16777215","color":"0","code":"52","code_ascii":"4","confidence":"97"},"characater":{"poly":{"p":{"x":"1123","y":"640"},"p":{"x":"1132","y":"640"},"p":{"x":"1131","y":"665"},"p":{"x":"1123","y":"665"}},"bkcolor":"16777215","color":"0","code":"53","code_ascii":"5","confidence":"97"},"characater":{"poly":{"p":{"x":"1134","y":"640"},"p":{"x":"1139","y":"640"},"p":{"x":"1139","y":"664"},"p":{"x":"1133","y":"664"}},"bkcolor":"16777215","color":"0","code":"49","code_ascii":"1","confidence":"77"},"characater":{"poly":{"p":{"x":"1154","y":"639"},"p":{"x":"1163","y":"639"},"p":{"x":"1163","y":"663"},"p":{"x":"1153","y":"663"}},"bkcolor":"16777215","color":"0","code":"66","code_ascii":"B","confidence":"97"},"characater":{"poly":{"p":{"x":"1164","y":"638"},"p":{"x":"1173","y":"638"},"p":{"x":"1173","y":"663"},"p":{"x":"1163","y":"663"}},"bkcolor":"16777215","color":"0","code":"66","code_ascii":"B","confidence":"94"},"characater":{"poly":{"p":{"x":"1191","y":"637"},"p":{"x":"1206","y":"636"},"p":{"x":"1205","y":"660"},"p":{"x":"1190","y":"661"}},"bkcolor":"16777215","color":"0","code":"76","code_ascii":"L","confidence":"34"}},"det_time_us":"1072592","poly":{"p":{"x":"1088","y":"642"},"p":{"x":"1210","y":"634"},"p":{"x":"1210","y":"661"},"p":{"x":"1087","y":"669"}}}},"det_time_us":"1720812"}}}

关联

输入文件:https ://drive.google.com/file/d/18wCzjMBpw7SIeVFByAGPQiqCBjg_0te3/view?usp=sharing

json data
  • 3 个回答
  • 867 Views
Martin Hope
neverMind9
Asked: 2018-05-27 02:08:48 +0800 CST

在不读取数据的情况下验证磁盘完整性?

  • 2

有一个适用于 Windows 的名为“HDDscan”的应用程序:http: //hddscan.com/

它还可以通过读取磁盘来验证磁盘,方法是它们只被传输到驱动器的内部缓冲区中,并且不知何故,只检查数据完整性。

来自网站:“在验证模式下,设备仅将数据块读取到驱动器的内部缓冲区并检查一致性,没有通过接口连接器/电缆传输数据。该程序测量每个块的操作时间。该程序从最小到最大逐个测试块。在读取模式下,设备读取数据块并将其通过接口传输到主机控制器。该程序将数据块读入临时缓冲区并测量每个块的操作时间。该程序从最小到最大逐个测试块。”

是否有可比较的方法来验证 Linux 中的数据完整性?

data integrity
  • 1 个回答
  • 169 Views

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