Question

我有这样的场景：

a = ["file1","file2","file3","file1","file2","file1","file5"]
set_flag = 0
for i in range (len(a)):
    file_name = '%s.txt' %(a[i])
    file_write = open('%s'%(file_name),'w')

以上工作和写入文件。但是，我想要包含一个逻辑，即使某些文件名在上面的列表中出现多次，file_write也只会发生一次。不应该有多个文件写入。例如。如果file1出现4次，它应该只写一次。有了这个，set_flag应设置为“1”，这样如果我尝试在我的代码中的任何地方写入file1，它应该绕过文件write.Any想法如何实现这一点并设置这样的标志..？

Answer 1

使用set()，它只存储项目的单个副本：

>>> a=["file1","file2","file3","file1","file2","file1","file5"]
>>> set(a)
set(['file3', 'file2', 'file1', 'file5'])

对于您的代码，这也将维护订单。：

a=["file1","file2","file3","file1","file2","file1","file5"]
seen=set()
for file in a:               #you can iterate over a list itself
    if file not in seen:     #if file is not in the set then write 
       file_name= '%s.txt' %(file)
       file_write= open('%s'%(file_name),'w')
       seen.add(file)  #add file to seen

Answer 2

一个集合是一个好主意，所以你只需遍历列表中的唯一文件名。另外，不要使用range(len(...))和其他一些清理：

a = ["file1","file2","file3","file1","file2","file1","file5"]
set_flag = 0
for file_name in set(a):
    file_write = open(file_name + '.txt', 'w')

Answer 3

如果出于某种原因需要创建输出文件的顺序，则可以使用OrderedDict（在2.7 +中）：

for fname in OrderedDict.fromkeys(a):
    with open(fname, 'w') as fout:
        pass

只写一次文件

3 个答案: