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将字典保存到CSV之前忽略一些键/值对

热度:23   发布时间:2023-06-13 16:53:04.0

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?

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。

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.  
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