Question

我觉得这个Python代码可以大大缩短，但我几乎总是倾向于回归编写C风格的布局。在您看来，缩短它的最佳方法是什么？可读性是一个奖励，而非要求。

def compfiles(file1, file2):
    linecnt = 0
    for line1 in open(file1):
        line1 = line1.strip()
        hit = False
        for line2 in open(file2):
            line2 = line2.strip()
            if line2 == line1:
                hit = True
                break
        if not hit:
            print("Miss: file %s contains '%s', but file %s does not!" % (file1, line1, file2))
        linecnt += 1
    print("%i lines compared between %s and %s." % (linecnt, file1, file2))

fn = ["file1.txt", "file2.txt"]
compfiles(fn[0], fn[1])
compfiles(fn[1], fn[0])

Answer 1

您的代码效率极低，因为您open循环内的第二个文件迭代第一个文件。只需将第二个文件读入一个列表（或者更好的是set，它会为您提供平均O(1)查询时间）并使用in运算符。此外，您的linecnt变量只计算file1中的行数 - 您只需将行读入列表并在此列表中调用len即可获得相同的数字：

def compfiles(file1, file2):
    lines1 = [l.strip() for l in open(file1).read().split("\n")]
    lines2 = set([l.strip() for l in open(file2).read().split("\n")])
    for line in lines1:
        if not line in lines2:
            print("Miss: file %s contains '%s', but file %s does not!" % (file1, line, file2))
    print("%i lines compared between %s and %s." % (len(lines1), file1, file2))

Answer 2

def compfiles(file1, file2):
    with open(file1) as fin:
        set1 = set(fin)
    with open(file2) as fin:
        set2 = set(fin)
    ... # do some set operations

如果文件有重复的行或顺序很重要，请迭代file1

def compfiles(file1, file2):
    with open(file2) as fin:
        set2 = set(fin)
    with open(file1) as fin:
        for i, line in enumerate(fin):
            if line not in set2:
                print("Miss: file %s contains '%s', but file %s does not!" % (file1, line1, file2))           
        print("%i lines compared between %s and %s." % (i+1, file1, file2))

缩短Python代码

2 个答案: