如何提取扩展名并保存它们而不重复？

Question

我有这个密码：

    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")

这正是我想要的，但问题是代码中有这样的重复：

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

Answer 1

使用口述。这里有一个例子。。。你知道吗

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

Answer 2

有更多可能的解决方案和方法来解决这个问题。你知道吗

大多数人（以及其他人）都同意使用dict是正确的方法。你知道吗

比如这里的史蒂文。：D个

有人会说set（）更方便、更自然，但我看到的和我自己做的大多数测试都表明，出于某种原因，使用dict（）稍微快一点。至于原因，没人知道。这也可能会因Python版本的不同而有所不同。你知道吗

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））。为了检查一个项目是否在一个列表中，迭代是在一个列表上执行的，在最坏的情况下，迭代次数会随着列表的增加而增加。你知道吗

为了进一步了解这个问题，我建议你检查一下相关的问题，特别是提到的问题。你知道吗

因此，我同意steveb的观点，并提出以下准则：

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了这个算法，它的性能非常好。你知道吗