Question

我有一个有序字典，其中包含同一个键的多个值。我需要遍历项目并对每个值执行不同的操作。

该词典如下：

OrderedDict([('1', ['file a','file1','file2','file3', 'Substringb']), ('2', ['file c', 'Substringd']), ('3', ['filed', 'Substringe']),('4', ['file f', 'Substringg']), ('5', ['file h', 'Substringi'])]

每个项目都有文件路径和子字符串。我无法控制从其他函数收到多少文件。我想打开它们中的每一个并查找子串。现在，当我在下面执行操作时，我得到了太多的值来解压缩：

下面是我使用的代码：当我提取值并尝试遍历each_file时，它不起作用 - 打印每个只打印文件路径的第一个字符[ex：'C'当文件路径是“ 'C：\ Users \ xxxxxx \ Documents \ something.txt“]。请帮我解决这个问题。 TIA。

for each_key, (each_file, each_substr) in d.iteritems():  #Giving too many values to unpack error since there are multiple files coming in[no control on how many to expect] 
    for each in each_file:
       print each
       with open(each_file) as f: ## This is giving error as the each is not working as I would expect it to.

Answer 1

您只需要1个循环即可提取所有数据。只需使用iteritems()并解压缩值元组即可获得所需的字符串

for each_key, (each_file, each_substr) in d.iteritems():
    print each_key, each_file, each_substr

Answer 2

我想出来了。

for each_key, each_value in d.iteritems():   
each_substring = each_value[1] #Since my substring is appended at the end and am removing it from the each_values list in the below step after saving it here
each_file = each_value[:-1] #Grabbing all files except the last one
for each in each_file:
   print each
   with open(each_file) as f: ## Now this works!!!

＆＃34;对于each.key，（each_file，each_substr）在d.iteritems（）中：＆＃34;如果我们不知道预期会有多少项目，那就不是非常有效了[

迭代有序字典python的多个值[对于相同的键]

2 个答案: