python - 如何删除字典列表中的元素

Question

我遇到一个问题，我有一个打开和关闭的企业的 json 文件。我需要指定开放企业的数量，这就是我这样做的原因。但它返回“无”。请注意，我必须使用函数。另外，我没有使用简单的计数器，因为我必须实际删除已关闭的业务，因为我必须对它们做更多的事情。这不是重复的，因为我尝试了其他帖子所说的内容，它给了我 0。

这是 json 文件条目的样子:

{
    "business_id":"1SWheh84yJXfytovILXOAQ",
    "name":"Arizona Biltmore Golf Club",
    "address":"2818 E Camino Acequia Drive",
    "city":"Phoenix",
    "state":"AZ",
    "postal_code":"85016",
    "latitude":33.5221425,
    "longitude":-112.0184807,
    "stars":3.0,
    "review_count":5,
    "is_open":0,
    "attributes":{
        "GoodForKids":"False"
    },
    "categories":"Golf, Active Life",
    "hours":null
}

import json
liste_businesses=[]
liste_open=[]
def number_entries(liste_businesses):
    with open ('yelp.txt') as file:
        for line in file:
            liste_businesses.append(json.loads((line)))
    return (len(liste_businesses))
def number_open(liste_businesses):
    for e in range (len(liste_businesses)):
        if 'is_open' not in liste[e]:
           liste_open=liste_businesses.remove(liste[e])
        if int(liste[e]['is_open'])==int(0):
            liste_open=liste_businesses.remove(liste[e])
print(number_open(liste_businesses))

Answer 1

除非您正在处理内存限制，否则迭代您的企业列表并选择开放的企业可能会更简单:

def load_businesses():
    businesses = []
    with open('yelp.txt') as file:
        for line in file:
            businesses.append(json.loads(line))

    #  More idiomatic to return a list than to modify global state
    return businesses


def get_open_businesses(businesses):
    # Make a new list rather than modifying the old one
    open_businesses = []
    for business in businesses:
        if business.get('is_open', '0') != '0':
            open_businesses.append(business)

    return open_businesses


businesses = load_businesses()
open_businesses = get_open_businesses(businesses)
print(len(open_businesses))

如果您想对开放业务使用列表理解:

[b for b in businesses if b.get('is_open', '0') != '0']