Question

在下面的代码中，我将使用split_message列表并查询API。如果没有结果，我的下一步是将列表的索引递减-1，所以我从列表中删除最后一项并重试搜索。如果再没有结果，我想重复这个过程，直到收到结果。

import requests
import json

message = "if i can\'t let it go out of my mind"

split_message = message.split()

length = len(split_message)

def decrementList():
    initial_request = requests.get('http://ws.spotify.com/search/1/track.json?q='+"%20"
                                   .join(split_message[:]))
    json_string = initial_request.content
    json_dict = json.loads(json_string)
    num_results = json_dict['info']['num_results']
    if num_results == 0:
        print "Sorry, no match!"
    if num_results > 0:
        print "Result found!"

decrementList()

实际上，我的下一步是采取这个：

requests.get('http://ws.spotify.com/search/1/track.json?q='+"%20"
                                       .join(split_message[:]))

并将其转换为：

requests.get('http://ws.spotify.com/search/1/track.json?q='+"%20"
                                       .join(split_message[:-1]))

然后这个：

requests.get('http://ws.spotify.com/search/1/track.json?q='+"%20"
                                       .join(split_message[:-2]))

重复直到我得到一场比赛。

我可以想到非常非pythonic的方法来做到这一点，但如果声明似乎太多了。那么对于我想要实现的目标，什么是最好和最pythonic的解决方案呢？

Answer 1

使用list.pop：

>>> message = "if i can\'t let it go out of my mind"
>>> split_message = message.split()
>>> split_message.pop()
'mind'
>>> split_message
['if', 'i', "can't", 'let', 'it', 'go', 'out', 'of', 'my']
>>> split_message.pop()
'my'
>>> split_message
['if', 'i', "can't", 'let', 'it', 'go', 'out', 'of']

Answer 2

修改

实际上，我更喜欢这个版本。有点清洁恕我直言。

def decrementList(words): words_remaining = words[:] while words_remaining: url = 'http://ws.spotify.com/search/1/track.json?q=' request = requests.get(url + '%20'.join(words_remaining)) json_dict = json.loads(request.content) if json_dict['info']['num_results'] > 0: num_words_taken = len(words) - len(words_remaining) return num_words_taken, json_dict words_remaining.pop()

从@falsetru获得list.pop灵感，请参阅其他答案。

原始回答

也许这更符合你的喜好：

import requests import json message = "if i can\'t let it go out of my mind" split_message = message.split() def decrementList(words): for w in [words] + [words[:-x] for x in range(1,len(words))]: url = 'http://ws.spotify.com/search/1/track.json?q=' request = requests.get(url + "%20".join(w)) json_dict = json.loads(request.content) num_results = json_dict['info']['num_results'] if num_results > 0: num_removed = len(words) - len(w) return num_removed, json_dict num_words_removed, json_dict = decrementList(split_message)

关键是[words] + [words[:-x] for x in range(1,len(words))]。假设words = message.split()，您将获得列表

[ ['if', 'i', 'can', 'let', 'it', 'go', 'out', 'of', 'my', 'mind'], ['if', 'i', 'can', 'let', 'it', 'go', 'out', 'of', 'my'], ['if', 'i', 'can', 'let', 'it', 'go', 'out', 'of'], ['if', 'i', 'can', 'let', 'it', 'go', 'out'], ['if', 'i', 'can', 'let', 'it', 'go'], ['if', 'i', 'can', 'let', 'it'], ['if', 'i', 'can', 'let'], ['if', 'i', 'can'], ['if', 'i'], ['if'] ]

我相信你做的是你想要的，它可以说是'Pythonic'。一旦得到想要的结果，该功能就会突破循环。它返回您必须从列表中删除的单词数以获得一些结果，以及请求产生的JSON字典。

如何在循环中减少列表索引

2 个答案: