Question

我有这段代码：

    x.write("FolderName\t\"%s\"\nPackName\t\"%s\"\n"% (pakedfolder, filename))
    x.write("\nList\t%s\n{\n" % (comptype))
    for root, dirs, files in os.walk("ymir work"):
       for file in files:
           file = path.splitext(file)[1]
           x.write("\t\"%s\"\n" %(file))
    x.write("}\n")
    x.write("\nList\tFilelist\n{\n")

生成我想要的，但问题是代码中有重复，如下所示：

".dds"
".mse"
".mse"
".mse"
".mse"
".mse"
".mse"
".dds"
".dds"
".dds"

Answer 1

有更多可能的解决方案和方法来解决这个问题。

大多数人（以及SO也同意）使用词典是正确的方法。

例如，这里以steveb为例。：d

有些人认为set（）会更方便自然，但是我看到的大多数测试都表明，由于某些原因，使用dict（）会稍快一些。至于为什么，没有人真正知道。此外，从Python版本到Python版本也可能不同。

字典和集合使用哈希来访问数据，这使得它们比列表更快（O（1））。要检查项是否在列表中，将对列表执行迭代，并且在最坏的情况下，迭代次数随列表一起增长。

要了解有关此主题的更多信息，我建议您检查相关问题，尤其是可能重复的问题。

所以，我同意steveb并提出以下代码：

chkdict = {} # A dictionary that we'll use to check for existance of an entry (whether is extension already processed or not)
setdef = chkdict.setdefault # Extracting a pointer of a method out of an instance may lead to faster access, thus improving performance a little
# Recurse through a directory:
for root, dirs, files in os.walk("ymir work"):
    # Loop through all files in currently examined directory:
    for file in files:
        ext = path.splitext(file) # Get an extension of a file
        # If file has no extension or file is named ".bashrc" or ".ds_store" for instance, then ignore it, otherwise write it to x:
        if ext[0] and ext[1]: ext = ext[1].lower()
        else: continue
        if not ext in chkdict:
            # D.setdefault(k[, d]) does: D.get(k, d), also set D[k] = d if k not in D
            # You decide whether to use my method with dict.setdefault(k, k)
            # Or you can write ext separately and then do: chkdict[ext] = None
            # Second solution might even be faster as setdefault() will check for existance again
            # But to be certain you should run the timeit test
            x.write("\t\"%s\"\n" % setdef(ext, ext))
            #x.write("\t\"%s\"\n" % ext)
            #chkdict[ext] = None
del chkdict # If you're not inside a function, better to free the memory as soon as you can (if you don't need the data stored there any longer)

我在大量数据上使用此算法，效果非常好。

Answer 2

使用词典。这是一个例子......

files = ['this.txt', 'that.txt', 'file.dat', 'a.dat', 'b.exe']

exts = {}

for file in files:
    exts[file.split('.')[1]] = 1

for ext in exts:
    print(ext)

输出：

dat
txt
exe

如何提取扩展并保存它们而不重复？

2 个答案: