Question

因此，我有一个包含嵌套字典的列表，而我想做的就是从字典中读取视图的值，然后将它们放入乘数。

video_ads=({'title': 'Healthy Living', 'company': 'Health Promotion 
Board','views': 15934},
{'title': 'Get a ride, anytime anywhere', 'company': 'Uber', 'views': 
923834},
{'title': 'Send money with GrabPay', 'company': 'Grab', 'views': 23466},
{'title': 'Ubereats now delivers nationwide', 'company': 'Uber', 'views': 
1337},
{'title': 'Get cabs now with UberFlash', 'company': 'Uber', 'views': 90234})

所以我想做的就是获得标题以及因某种因素而产生的多种观点，这就是我到目前为止所写的内容

def payment_for_views(x):
    for x in range(0,5):
        x = x+1

        if video_ads[x]['views'] >= 50000:
            payment = video_ads[x]['views']*0.55
        elif views <50000 and video_ads[x]['views'] >=10000:
            payment=video_ads[x]['views']*0.68
        else:
            payment = video_ads[x]['views']*0.82
    #print(x)
    print(video_ads[x]['title'] + ': $'+ str(payment))
    print(video_ads[x]['title'] + ': $'+ str(payment))
    print(video_ads[x]['title'] + ': $'+ str(payment))
    print(video_ads[x]['title'] + ': $'+ str(payment))
    print(video_ads[x]['title'] + ': $'+ str(payment))

例如，似乎只打印一行：

 Get cabs now with UberFlash: $49628.700000000004
 Get cabs now with UberFlash: $49628.700000000004
 Get cabs now with UberFlash: $49628.700000000004
 Get cabs now with UberFlash: $49628.700000000004
 Get cabs now with UberFlash: $49628.700000000004

这就是我所得到的。我需要5行所有的标题和数量。救命！！！并预先感谢！

Answer 1

video_ads=({'title': 'Healthy Living', 'company': 'Health Promotion Board','views': 15934},
           {'title': 'Get a ride, anytime anywhere', 'company': 'Uber', 'views': 923834},
           {'title': 'Send money with GrabPay', 'company': 'Grab', 'views': 23466},
           {'title': 'Ubereats now delivers nationwide', 'company': 'Uber', 'views': 1337},
           {'title': 'Get cabs now with UberFlash', 'company': 'Uber', 'views': 90234})

def payment_for_views(item):
    if item >= 50000:
        payment = item * 0.55
    elif item < 50000 and item >= 10000:
        payment = item * 0.68
    else:
        payment = item * 0.82
    return payment

[print(per_ad['title'], ': $' + str(payment_for_views(per_ad['views']))) for per_ad in video_ads]

输出：

健康生活：$ 10835.12

随时随地乘车：$ 508108.70000000007

使用GrabPay汇款：$ 15956.880000000001

Ubereats现在在全国范围内销售：1096.34美元

立即使用UberFlash出租车：$ 49628.700000000004

Answer 2

for a in video_ads:
    print('%s: $%.2f' % (a['title'], a['views'] * (0.55 if a['views'] >= 50000 else 0.68 if a['views'] >= 10000 else 0.82)))

这将输出：

Healthy Living: $10835.12
Get a ride, anytime anywhere: $508108.70
Send money with GrabPay: $15956.88
Ubereats now delivers nationwide: $1096.34
Get cabs now with UberFlash: $49628.70

Answer 3

您需要将print语句放入for循环中。

在循环的前后留置print意味着x将固定为for循环中设置的最终值。

def payment_for_views(x):
    for x in range(len(x)):
        if video_ads[x]['views'] >= 50000:
            payment = video_ads[x]['views']*0.55
        elif video_ads[x]['views'] <50000 and video_ads[x]['views'] >=10000:
            payment=video_ads[x]['views']*0.68
        else:
            payment = video_ads[x]['views']*0.82
        print(video_ads[x]['title'] + ': $'+ str(payment))

有多种方法可以改善代码：

直接迭代字典，而不是按索引进行迭代。
为每个字典提取一次“ views”值，然后分配给一个变量。
使用链式比较来检查一个值是否在其他两个值之间。
使用指定了十进制精度的f字符串（Python 3.6 +）。

这是一个演示：

def payment_for_views(x):
    for item in x:
        views = item['views']
        if views >= 50000:
            factor = 0.55
        elif 10000 <= views < 50000:
            factor = 0.68
        else:
            factor = 0.82

        print(f'{item["title"]}: ${views*factor:.2f}')

payment_for_views(video_ads)

结果：

Healthy Living: $10835.12
Get a ride, anytime anywhere: $508108.70
Send money with GrabPay: $15956.88
Ubereats now delivers nationwide: $1096.34
Get cabs now with UberFlash: $49628.70

如何遍历带有嵌套字典的列表？

3 个答案: