问题描述
我是Python新手（2周），陷入一个问题。
我有这本不断增长的字典：
dict1={"some_large_key1":"value1","some_large_key2":"value2",........,"some_large_key1000":"value1000"}
问题1：我想创建一个csv文件，我不想在其中包含一些键，值对。
->一种可能的解决方案是创建一个键列表，而忽略整个列表。
list1=["some_large_key1","some_large_key2","some_large_key3"]
for key,value in dict1:
if key not in list1: #something like this
#do something
else:
#do something
问题2：我不想给出这么大的键名，相反，我只能给出：
list1=["key1","key2",...]
是正确的方法还是我应该考虑其他方法？
1楼
您的“ some_large_key1”和“ key1”之间的关系不是很清楚。 但是，我认为您可以执行以下操作：
original_dict = {
"some_large_key1": "value1",
"some_large_key2": "value2",
......,
"some_large_key1000": "value1000"
}
to_ignore = ['key1', 'key2', 'key3', ...]
filtered_dict = {key: value for (key, value) in original_dict.items() if key.rsplit('_', 1)[-1] not in to_ignore}
key.rsplit('_', 1)[-1]作用是删除了some_large_前缀；
您可以用将长键名转换成短键名的功能来代替此部分。
2楼
是的，您可以列出要忽略的键。
dict1={"some_large_key1":"value1","some_large_key2":"value2","some_large_key3":"value3"}
ignore_list=['key1', 'key2']
new_dict = {}
for key in dict1:
if not any(x in key for x in ignore_list):
key_list=[key]
value_list=[dict1[key]]
#make a new dict from above 2 lists (This will exclude ignore key,value pair)
for i in range(len(key_list)):
new_dict[key_list[i]] = value_list[i]
print new_dict
OUTPUT：
{'some_new_key3':'value3'} #Note that key1,key2 is ignored in this output dictionary.