使用 python difflib 比較兩個以上的文件

Question

我想通過將它們相互比較並突出顯示差異來概述例如多台（3+）計算機的 ldd 依賴項列表。 例如，如果我有一個如下所示的 dict：

my_ldd_outputs = {
  01:"<ldd_output>",
  02:"<ldd_output>", 
  ...
  09:"<ldd_output>",
  10:"<ldd_output>"
}

我希望 output 看起來像

<identical line 1>
<identical line 2>
<identical line 3>
<differing line 4> (computer 01 02)
<differing line 4> (computer 04 05 06 07)
<differing line 4> (computer 08 09 10)
<identical line 5>
<identical line 6>
...

我的第一種方法涉及python difflib ，我的想法是首先獲取一個數據結構，其中來自上述my_ldd_outputs字典的所有ldd_output列表（只是用\n拆分的結果）長度相同，並且任何缺失的行都存在於另一個ldd_output string 加上一個字符串。 因此，如果兩個文件看起來像這樣：

ldd_1 = """
<identical line 1>
<identical line 2>
<differing line 3>
<identical line 4>
<extra line 5>
<identical line 6>
"""

ldd_2 = """
<identical line 1>
<identical line 2>
<differing line 3>
<identical line 4>
<identical line 6>
"""

我的目標是將這些文件存儲為

ldd_1 = """
<identical line 1>
<identical line 2>
<differing line 3>
<identical line 4>
<extra line 5>
<identical line 6>
"""

ldd_2 = """
<identical line 1>
<identical line 2>
<differing line 3>
<identical line 4>
<None>
<identical line 6>
"""

最終只需遍歷轉換文件的每一行（現在都具有相同的長度）並比較每一行的差異並忽略任何<None>條目，以便可以連續打印差異。

我創建了一個 function ，它使用 python difflib用<None>字符串填充其他文件中缺少的行。 但是，我不確定如何擴展此 function 以包含任意數量的差異

def generate_diff(file_1, file_2):
    #differing hashvalues from ldd can be ignored, we only care about version and path
    def remove_hashvalues(input):
        return re.sub("([a-zA-Z0-9_.-]{32}\/|\([a-zA-Z0-9_.-]*\))", "<>", input)
    diff = [line.strip() for line in difflib.ndiff(remove_hashvalues(base).splitlines(keepends=True),remove_hashvalues(file_2).splitlines(keepends=True))]
    list_1 = []
    list_2 = []
    i = 0
    while i<len(diff):
        if diff[i].strip():
            if diff[i][0:2]=="- ":
                lost = []
                gained = []
                while diff[i][0:2]=="- " or diff[i][0:2]=="? ":
                    if diff[i][0:2]=="- ": lost.append(diff[i][1:].strip())
                    i+=1
                while diff[i][0:2]=="+ " or diff[i][0:2]=="? ":
                    if diff[i][0:2]=="+ ": gained.append(diff[i][1:].strip())
                    i+=1
                while len(lost) != len(gained):
                    lost.append("<None>") if len(lost)<len(gained) else gained.insert(0,"<None>")
                list_1+=lost; list_2+=gained
            elif diff[i][0:2]=="+ ":
                list_1.append("<None>"); list_2.append(diff[i][1:].strip())
            if not diff[i][0:2]=="? ":
                list_1.append(diff[i].strip()); list_2.append(diff[i].strip())
        i+=1
    return list_1, list_2

我還發現 這個工具可以比較多個文件，但不幸的是它不是用來比較代碼的。

編輯：我調整了@AyoubKaanich 的解決方案建議，以創建一個更簡化的版本來滿足我的需求：

from collections import defaultdict
import re
def transform(input):
    input = re.sub("([a-zA-Z0-9_.-]{32}\/|\([a-zA-Z0-9_.-]*\))", "<>", input) # differing hashvalues can be ignored, we only care about version and path
    return sorted(input.splitlines())
def generate_diff(outputs: dict):
    mapping = defaultdict(set)
    for target, output in outputs.items():
        for line in transform(output):
            mapping[line.strip()].add(target)
    result = []
    current_line = None
    color_index = 0
    for line in sorted(mapping.keys()):
        if len(outputs) == len(mapping[line]):
            if current_line: current_line = None
            result.append((line))
        else:
            if current_line != line.split(" ")[0]:
                current_line = line.split(" ")[0]
                color_index+=1
            result.append((f"\033[3{color_index%6+1}m{line}\033[0m",mapping[line]))
    return result

唯一的缺點是這不適用於字符串在任意部分變化的差異，這是difflib擅長檢測的。 但是，對於ldd的情況，由於總是首先列出依賴項，因此按字母順序排序並獲取字符串的第一部分是可行的。

Answer 1

git 與 Git 不同。

git add ldd.tmp1 ldd.tmp2 ldd.tmp3
git diff -- | grep + > result.tmp

見https://git-scm.com/docs/git-diff

還：

• 融合
• Kdiff3

Answer 2

純 Python 解決方案，沒有庫或額外的依賴項。

注意：由於一些假設，此解決方案有效：

• 行的順序無關緊要
• 一條線存在或缺失（沒有邏輯來檢查線之間的相似性）

from collections import defaultdict
import re

def transform(input):
    # differing hashvalues from ldd can be ignored, we only care about version and path
    input = re.sub("([a-zA-Z0-9_.-]{32}\/|\([a-zA-Z0-9_.-]*\))", "<>", input)
    return sorted(input.splitlines())

def generate_diff(outputs: dict, common_threshold = 0):
    """
        common_threshold: how many outputs need to contain line to consider it common
            and mark outputs that do not have it as missing
    """
    assert(common_threshold <= len(outputs))

    mapping = defaultdict(set)
    for target, output in outputs.items():
        for line in transform(output):
            mapping[line].add(target)
    
    for line in sorted(mapping.keys()):
        found = mapping[line]
        if len(outputs) == len(found):
            print('  ' + line)
        elif len(found) >= common_threshold:
            missed_str = ",".join(map(str, set(outputs.keys()) - found))
            print(f'- {line}  ({missed_str})')
        else:
            added_str = ",".join(map(str, found))
            print(f'+ {line}  ({added_str})')

樣品執行

my_ldd_outputs = {
'A': """
linux-vdso.so.1 (0x00007ffde4f09000)
libtinfo.so.6 => /lib/x86_64-linux-gnu/libtinfo.so.6 (0x00007fe0594f3000)
libc.so.6 => /lib/x86_64-linux-gnu/libc.so.6 (0x00007fe0592cb000)
/lib64/ld-linux-x86-64.so.2 (0x00007fe059690000)
""",
'B': """
linux-vdso.so.1 (0x00007fff697b6000)
libc.so.6 => /lib/x86_64-linux-gnu/libc.so.6 (0x00007f1c54045000)
/lib64/ld-linux-x86-64.so.2 (0x00007f1c54299000)
""",
'C': """
linux-vdso.so.1 (0x00007fffd61f9000)
libcrypto.so.3 => /lib/x86_64-linux-gnu/libcrypto.so.3 (0x00007f08a51a3000)
libc.so.6 => /lib/x86_64-linux-gnu/libc.so.6 (0x00007f08a4f7b000)
/lib64/ld-linux-x86-64.so.2 (0x00007f08a5612000)
""",
'D': """
linux-vdso.so.1 (0x00007ffcf9ddd000)
libcrypto.so.3 => /lib/x86_64-linux-gnu/libcrypto.so.3 (0x00007fa2e381b000)
libselinux.so.1 => /lib/x86_64-linux-gnu/libselinux.so.1 (0x00007fa2e37ef000)
libc.so.6 => /lib/x86_64-linux-gnu/libc.so.6 (0x00007fa2e35c7000)
libpcre2-8.so.0 => /lib/x86_64-linux-gnu/libpcre2-8.so.0 (0x00007fa2e3530000)
/lib64/ld-linux-x86-64.so.2 (0x00007fa2e3cd7000)
""",
'E': """
linux-vdso.so.1 (0x00007ffc2deab000)
libcrypto.so.3 => /lib/x86_64-linux-gnu/libcrypto.so.3 (0x00007f31fed91000)
libz.so.1 => /lib/x86_64-linux-gnu/libz.so.1 (0x00007f31fed75000)
libselinux.so.1 => /lib/x86_64-linux-gnu/libselinux.so.1 (0x00007f31fed49000)
libgssapi_krb5.so.2 => /lib/x86_64-linux-gnu/libgssapi_krb5.so.2 (0x00007f31fecf5000)
libc.so.6 => /lib/x86_64-linux-gnu/libc.so.6 (0x00007f31feacd000)
libpcre2-8.so.0 => /lib/x86_64-linux-gnu/libpcre2-8.so.0 (0x00007f31fea34000)
/lib64/ld-linux-x86-64.so.2 (0x00007f31ff2af000)
libkrb5.so.3 => /lib/x86_64-linux-gnu/libkrb5.so.3 (0x00007f31fe969000)
libk5crypto.so.3 => /lib/x86_64-linux-gnu/libk5crypto.so.3 (0x00007f31fe93a000)
libcom_err.so.2 => /lib/x86_64-linux-gnu/libcom_err.so.2 (0x00007f31fe934000)
libkrb5support.so.0 => /lib/x86_64-linux-gnu/libkrb5support.so.0 (0x00007f31fe926000)
libkeyutils.so.1 => /lib/x86_64-linux-gnu/libkeyutils.so.1 (0x00007f31fe91f000)
libresolv.so.2 => /lib/x86_64-linux-gnu/libresolv.so.2 (0x00007f31fe909000)
"""
}
generate_diff(my_ldd_outputs, 2)

輸出

  /lib64/ld-linux-x86-64.so.2 <>
  libc.so.6 => /lib/x86_64-linux-gnu/libc.so.6 <>
+ libcom_err.so.2 => /lib/x86_64-linux-gnu/libcom_err.so.2 <>  (E)
- libcrypto.so.3 => /lib/x86_64-linux-gnu/libcrypto.so.3 <>  (B,A)
+ libgssapi_krb5.so.2 => /lib/x86_64-linux-gnu/libgssapi_krb5.so.2 <>  (E)
+ libk5crypto.so.3 => /lib/x86_64-linux-gnu/libk5crypto.so.3 <>  (E)
+ libkeyutils.so.1 => /lib/x86_64-linux-gnu/libkeyutils.so.1 <>  (E)
+ libkrb5.so.3 => /lib/x86_64-linux-gnu/libkrb5.so.3 <>  (E)
+ libkrb5support.so.0 => /lib/x86_64-linux-gnu/libkrb5support.so.0 <>  (E)
- libpcre2-8.so.0 => /lib/x86_64-linux-gnu/libpcre2-8.so.0 <>  (C,B,A)
+ libresolv.so.2 => /lib/x86_64-linux-gnu/libresolv.so.2 <>  (E)
- libselinux.so.1 => /lib/x86_64-linux-gnu/libselinux.so.1 <>  (C,B,A)
+ libtinfo.so.6 => /lib/x86_64-linux-gnu/libtinfo.so.6 <>  (A)
+ libz.so.1 => /lib/x86_64-linux-gnu/libz.so.1 <>  (E)
  linux-vdso.so.1 <>