比较 2 个分隔文件并输出差异

比较 2 个分隔文件并输出差异

过去我曾多次在这里提出类似的问题,并取得了巨大的成功,但现在我的需求略有变化,我正在努力获得我正在寻找的确切输出。

我想比较2相似的分隔文件,但它们将具有不同的行数和一些重复项。这些文件将具有相同的标头。

file1.txt

mem_id     date     time     building
aa1         bb1      cc1     dd1
aa2         bb2      cc2     dd2
aa3         bb3      ccx3    dd3
aa4         bb4      cc4     dd4
aa5         bb5      cc5     dd5

file2.txt

mem_id     date     time     building
aa1         bby1     cc1     ddy1
aa2         bb2      cc2     dd2  
aa3         bb3      cc3     dd3
aa4         bb4      cc4     dd4
aa4         bb4a     cc4a    dd4a

你会发现有 4 个不同点:

1- File2,mem_id aa1 在“日期”和“建筑物”列中都有“y”

2- File1,mem_id aa3 在“时间”列中有一个“x”

3- File1,有一个 mem_id aa5

4- File2,mem_id aa4 有 2 个条目

我想运行一个脚本来仅输出两个文件之间的差异(跳过相同的行)。我尝试过的所有内容都会挂在重复或跳过的行上,从而弄乱整个文件的输出。如果所有行都匹配,则以下代码可以正常运行:

current_code

awk -F ',' 'BEGIN {IGNORECASE = 1} NR==1 {for (i=1; i<=NF; i++)    header[i] = $i}NR==FNR {for (i=1; i<=NF; i++) {A[i,NR] = $i} next}{  for (i=1; i<=NF; i++) if (A[i,FNR] != $i) print header[1]"#-"$1": " header[i] "- " ARGV[1] " value= ", A[i,FNR]" / " ARGV[2] " value= "$i}'

desired_output.txt

Mem_id#-aa1 : date-  file1.txt value = bb1 / file2.txt value= bby1
Mem_id#-aa1 : building-  file1.txt value = dd1 / file2.txt value= ddy1
Mem_id#-aa3 : time-  file1.txt value = ccx3 / file2.txt value= dd3
Mem_id#-aa4 : date-  file1.txt value =        / file2.txt value= bb4a
Mem_id#-aa4 : time-  file1.txt value =        / file2.txt value= cc4a
Mem_id#-aa4 : building-  file1.txt value =        / file2.txt value= dd4a
Mem_id#-aa5 : date-  file1.txt value = bb5 / file2.txt value= 
Mem_id#-aa5 : time-  file1.txt value =  cc5 / file2.txt value= 
Mem_id#-aa5 : building-  file1.txt value =  dd5 / file2.txt value= 

答案1

以下 python 程序应该执行您想要的操作,或者非常接近它的操作。

  • 第三行desired_output.txt似乎是错误的:

    Mem_id#-aa3 : time-  file1.txt value = ccx3 / file2.txt value= dd3
    

    cc3`dd3 should probably be

    除此之外,程序的输出除了空格之外都匹配,这在示例输出中似乎有点不规则。

  • 输入被认为是按键(memid)排序的

  • 程序在尝试同步时默认缓冲 4 行 (max_diff + 1)。如果该缓冲区中的键中没有一个与“当前”键匹配,则 vv 均被视为不匹配并打印,并尝试下一对。如果找到某个键​​,则将其他缓冲区中的不匹配项或首先输出。
  • 当第一行和第二行具有相同的 memid 两次(或更多)时,示例输入对预期的行为有一些限制。

    output()我尝试匹配任何行并弹出所有匹配项(从左到右)。因此,同一 memid 中匹配行的顺序并不重要。如果左侧或右侧或两者都为空,则打印很容易(特别是当两者都为空时)。对于其余的,我从左到右匹配剩余的每一行。

  • fmt中的字符串决定line_out()输出,您可以自由更改/重新排序。

#! /usr/bin/env python
# coding: utf-8
# http://unix.stackexchange.com/q/161913/33055

from __future__ import print_function
from collections import OrderedDict
from logging import debug

import sys


class RowBuffer:
    def __init__(self, file_name, delim=None, max_diff=3):
        """delim is the character that is used for splitting input.
        None->whitespace
        """
        self._verbose = 0
        self._file_name = file_name
        self._fp = open(self._file_name)
        self._delim = delim
        self._max_diff = max_diff
        self._head = self._fp.readline().split(delim)
        # the buffer consists of a maximum of max_diff entries
        # the keys are the first items of a row, the value a list
        # of all other items on that row
        self._buffer = OrderedDict()
        self.fill_buffer()

    def compare(self, rb):
        """check if self._buffer"""
        if self._head != rb._head:
            print('headings differ:\n  {}\n  {}'.format(
                self._head, rb._head))
        while self._buffer:
            l = self.get()
            try:
                r = rb.get()
            except KeyError:
                debug('only left %s', l[0])
                self.output(l, None, rb)
                break
            if l[0] == r[0]:
                debug('compare vals %s', l[0])
                self.output(l, r, rb)
                continue
            if l[0] in rb:
                # left key in right, but not at top
                # output right until top keys are same
                while l[0] != r[0]:
                    debug('only right %s', r[0])
                    self.output(None, r, rb)
                    r = rb.get()
                self.output(l, r, rb)
                continue
            if r[0] in self:
                # right key in left, but not at top
                # output left until top keys are same
                while l[0] != r[0]:
                    debug('only left %s', l[0])
                    self.output(l, None, rb)
                    l = self.get()
                self.output(l, r, rb)
                continue
            # neither found: output both
            debug('neither left in right nor vv %s %s', l[0], r[0])
            self.output(l, None, rb)
            self.output(None, r, rb)
        while rb._buffer:  # remaining in right file
            r = rb.get()
            debug('only right %s', r[0])
            self.output(None, r, rb)

    def output(self, l, r, right):
        fmt1 = '{col0_header}#-{col0_value} : {col_header}-  ' \
            '{left_file_name} value = {left_value} / ' \
            '{right_file_name} value= {right_value}'
        d = dict(
            col0_header=self._head[0],
            left_file_name=self._file_name,
            right_file_name=right._file_name,
        )
        if l is not None and r is not None:
            # one or more values on both sides, compare all lines on the
            # left with all on the right remove any matching pairs
            match = {}  # left index to right index
            for lidx, lv in enumerate(l[1]):
                for ridx, rv in enumerate(r[1]):
                    if lv == rv:
                        if lidx not in match:
                            match[lidx] = ridx
            # pop from back of list, not invalidate index
            for lidx in sorted(match, reverse=True):
                l[1].pop(lidx)
            for ridx in sorted(match.values(), reverse=True):
                r[1].pop(lidx)
        if r is None or not r[1]:
            for lv in l[1]:
                for idx, k in enumerate(self._head[1:]):
                    self.line_out(d, col0_value=l[0], col_header=k,
                                  left_value=lv[idx], right_value='    ')
            return
        if l is None or not l[1]:
            for rv in r[1]:
                for idx, k in enumerate(self._head[1:]):
                    self.line_out(d, col0_value=l[0], col_header=k,
                                  left_value='    ', right_value=rv[idx])
            return
        # print non matching
        for lv in l[1]:
            for rv in r[1]:
                for idx, k in enumerate(self._head[1:]):
                    if lv[idx] == rv[idx]:
                        continue  # same value
                    self.line_out(d, col0_value=l[0], col_header=k,
                                  left_value=lv[idx], right_value=rv[idx])

    def line_out(self, d, **kw):
        # manipulate and print output
        # the fields of the format string can be arbitrarily arranged
        # as long as the field names (between {} match)
        fmt = '{col0_header}#-{col0_value} : {col_header}-  ' \
            '{left_file_name} value = {left_value} / ' \
            '{right_file_name} value= {right_value}'
        d1 = d.copy()
        d1.update(kw)
        s = fmt.format(**d1)
        # s = s.rstrip()
        s = s[0].upper() + s[1:]  # sample output doesn't match input
        print(s)

    def get(self):
        item = self._buffer.popitem(last=False)
        self.fill_buffer()
        return item

    def fill_buffer(self):
        if self._fp is None:
            return
        while len(self._buffer) < self._max_diff:
            row = self._fp.readline().split(self._delim)
            if not row:
                self._fp.close()
                self._fp = None
                return
            entry = self._buffer.setdefault(row[0], [])
            entry.append(row[1:])

    def __contains__(self, key):
        self.fill_buffer()
        return key in self._buffer

rb1 = RowBuffer(sys.argv[1])
rb2 = RowBuffer(sys.argv[2])
rb1.compare(rb2)

答案2

这是对您的问题的(远非优雅的)部分解决方案。它使用第一列作为 id 列(它不必是第一列,但你绝对必须有一个)并引入第三个维度suffix来存储同一键的多次出现。最后它尝试找到文件 2 中那些在文件 1 中没有找到的键。

BEGIN {
    IGNORECASE = 1
} 

NR==1 {
    for (i = 1; i <= NF; i++)    
        header[i] = $i
    suffix = 0
    previous_key=""
}

NR==FNR {
    if ($1 == previous_key) {
        suffix = suffix + 1
        max_suffix[$1] = suffix
    } else
        suffix = 0
    for (i = 1; i <= NF; i++) {
        A[$1,suffix,i] = $i
    } 
    key_count[$1] = key_count[$1] + 1
    previous_key = $1
    next
}

{  
    if ($1 == previous_key)
        suffix = suffix + 1
    else
        suffix = 0
    previous_key = $1
    if (A[$1,suffix,1] != "") {
        for (i = 2; i <= NF; i++) 
            if (A[$1,suffix,i] != $i) {
                print header[1]"#-"$1": " header[i] "- " ARGV[1] " value= ", A[$1,suffix,i]" / " ARGV[2] " value= "$i
            }
        key_count[$1] = key_count[$1] - 1
    }
    else
        for (i = 2; i <= NF; i++) 
            print header[1]"#-"$1": " header[i] "- " ARGV[1] " value= ", " / " ARGV[2] " value= "$i
}

END {
    for (missing_key in key_count) 
        if (key_count[missing_key] > 0) {            
            for (suffix = max_suffix[missing_key] - key_count[missing_key] + 1; suffix <= C[missing_key]; suffix++) 
                for (i = 2; i <= NF; i++) 
                    print header[1]"#-"missing_key": " header[i] "- " ARGV[1] " value= ", A[missing_key,suffix,i] " / " ARGV[2] " value= "
        }           
}

有一个警告:文件 2 中的不匹配条目始终打印在末尾,并且不根据文件中的位置进行排序。此外,这些行的排序是任意的。我想这可以通过将结果传递到命令中来实现sort

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