Question

如果其中一个键中包含字符，如何删除JSON中具有的列表的字典：

for i in data:
    results = i["results"]
    if not results == []:
        for x in results:
        price_str = x["price_str"]
        if "await" in price_str:
            results.remove(x)

我的输入：

"results": [{
  "price_str": "results awaited",
  "marque": "samsung"
  },
  {
  "price_str": "sold",
  "marque": "apple"
  }]

我想要的输出：

"results":[{
  "price_str": "sold",
  "marque": "apple"
  }]

Answer 1

在迭代时从列表中删除元素的正确方法是在列表的副本上进行迭代：这样一来，您不会得到意想不到的结果，因为您没有在编辑要迭代的列表。

data = {
    "results": [{
            "price_str": "results awaited",
            "marque": "samsung"
        }, {
            "price_str": "sold",
            "marque": "apple"
        }
    ]
}

for results in data.itervalues():

    # You don't need to check if the list is empty
    # The for loop doesn't start if the list is empty
    # if not results == []:

    # Iterates over a copy of the list. So when you modify the original
    # list, you do not modify the copy that you iterate over.
    for result in results[:]:
        price_str = result["price_str"]
        if "await" in price_str:
            results.remove(result)

print(data)

Answer 2

尝试一下：

dct = {'results' : [{
  "price_str": "results awaited",
  "marque": "samsung"
  },
  {
  "price_str": "sold",
  "marque": "apple"
  }]}

for entry, x in enumerate(dct['results']):
    if 'await' in x['price_str']:
        print(x)
        dct['results'].pop(entry)

Answer 3

您也可以尝试

dct = {'results' : [{
  "price_str": "results awaited",
  "marque": "samsung"
  },
  {
  "price_str": "sold",
  "marque": "apple"
  }]}


items = dct['results']
y=[a for a in items if "await" not in a['price_str']]

y将具有过滤的项目列表。然后，以后可以根据需要将其重新分配给词典。

Answer 4

具有此初始指示：

list = {
            'results': [
                {'marque': 'samsung', 'price_str': 'results awaited'},
                {'marque': 'apple', 'price_str': 'sold'}
            ]
       }

通过使用filter()，您可以从列表中获取适合您条件的元素。在您的情况下，您还可以使用find()函数来获得不适合您搜索的结果，因此您将获得所需的数组：

>>> new_results = filter(lambda x: x['price_str'].lower().find("await") < 0, list["results"])
>>> new_results
[{'marque': 'apple', 'price_str': 'sold'}]
>>> list["results"] = new_results
>>> list
{'results': [{'marque': 'apple', 'price_str': 'sold'}]}

希望有帮助！

如果键值中包含字符串，则删除字典

4 个答案: