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嘿伙计们，非常感谢你花时间来看看我的问题，我已经在这个代码上工作了大约 1 周（我对编码和 python 也是 1 周的新手）目前只有在 xrange(x) 中的 x 时循环才有效和 'rp' : 'x' 是此 xml 中可用的正确行数。xml 全天更新，我想知道是否有人可以提供使 x 动态的解决方案？

import mechanize
import urllib
import json
import re
from sched import scheduler
from time import time, sleep

from sched import scheduler
from time import time, sleep

s = scheduler(time, sleep)

def run_periodically(start, end, interval, func):
event_time = start
while event_time < end:
    s.enterabs(event_time, 0, func, ())
    event_time += interval
s.run()

def getData():  
post_url = "urlofinterest_xml"
browser = mechanize.Browser()
browser.set_handle_robots(False)
browser.addheaders = [('User-agent', 'Firefox')]

######These are the parameters you've got from checking with the aforementioned tools
parameters = {'page' : '1',
              'rp' : '8',
              'sortname' : 'roi',
              'sortorder' : 'desc'
             }
#####Encode the parameters
data = urllib.urlencode(parameters)
trans_array = browser.open(post_url,data).read().decode('UTF-8')

xmlload1 = json.loads(trans_array)
pattern1 = re.compile('>&nbsp;&nbsp;(.*)<')
pattern2 = re.compile('/control/profile/view/(.*)\' title=')
pattern3 = re.compile('<span style=\'font-size:12px;\'>(.*)<\/span>')
pattern4 = re.compile('title=\'Naps posted: (.*) Winners:')
pattern5 = re.compile('Winners: (.*)\'><img src=')


for i in xrange(8):
    user_delimiter = xmlload1['rows'][i]['cell']['username']
    selection_delimiter = xmlload1['rows'][i]['cell']['race_horse']

    username_delimiter_results = re.findall(pattern1, user_delimiter)[0]
    userid_delimiter_results = int(re.findall(pattern2, user_delimiter)[0])
    user_selection = re.findall(pattern3, selection_delimiter)[0]
    user_numberofselections = float(re.findall(pattern4, user_delimiter)[0])
    user_numberofwinners = float(re.findall(pattern5, user_delimiter)[0])

    strikeratecalc1 = user_numberofwinners/user_numberofselections
    strikeratecalc2 = strikeratecalc1*100

    print "user id = ",userid_delimiter_results
    print "username = ",username_delimiter_results
    print "user selection = ",user_selection
    print "best price available as decimal = ",xmlload1['rows'][i]['cell']     ['tws.best_price']
    print "race time = ",xmlload1['rows'][i]['cell']['race_time']
    print "race meeting = ",xmlload1['rows'][i]['cell']['race_meeting']
    print "ROI = ",xmlload1['rows'][i]['cell']['roi']
    print "number of selections = ",user_numberofselections
    print "number of winners = ",user_numberofwinners
    print "Strike rate = ",strikeratecalc2,"%"
    print ""


getData()


run_periodically(time()+5, time()+1000000, 15, getData)

亲切的问候 AEA

4

2 回答 2

4

首先，我将讨论如何迭代结果。根据您的代码，xmlload1['rows']是一个字典数组，因此您可以直接迭代它，而不是选择任意数字。为了使它成为一个更好的例子，我将设置一些任意数据来说明这一点：

xmlload1 = {
   "rows": [{"cell": {"username": "one", "race_horse":"b"}}, {"cell": {"username": "two", "race_horse": "c"}}]
}

因此，鉴于上面的数据，您可以在 for 循环中遍历行，如下所示：

for row in xmlload1['rows']:
    cell = row["cell"]
    print cell["username"]
    print cell["race_horse"]

每次迭代，单元格都会获取可迭代对象（中的列表xmlload1['rows']）中另一个元素的值。这适用于任何支持迭代的容器或序列（如lists、tuples、dicts、generators等）

请注意我没有在任何地方使用任何幻数，所以xmlload1['rows']可以任意长并且它仍然可以工作。

您可以使用函数将请求设置为动态，如下所示：

def get_data(rp=8, page=1):
    parameters = {'page' : str(page),
              'rp' : str(rp),
              'sortname' : 'roi',
              'sortorder' : 'desc'
             }
    data = urllib.urlencode(parameters)
    trans_array = browser.open(post_url,data).read().decode('UTF-8')
    return json.loads(trans_array)

现在，您可以调用get_data(rp=5)来获取 5 行，或者get_data(rp=8)获取 8 行[并get_data(rp=8, page=3)获取第三页]等。您可以清楚地添加其他变量，甚至可以parameters直接将 dict 传递给函数。

于 2013-06-02T05:09:14.510 回答
2

我不确定我是否理解您的问题，但我认为您想要的是：

rows = xmlload1['rows']
for row in rows:
    user_delimiter = row['cell']['username']
    selection_delimiter = row['cell']['race_horse']
    # ...

如果您需要行索引以及行本身，请使用enumerate：

rows = xmlload1['rows']
for i, row in enumerate(rows):
    user_delimiter = row['cell']['username']
    selection_delimiter = row['cell']['race_horse']
    # ...

一般来说，如果你做for i in range(…)的不是固定数量的迭代，那么你做错了。通常有一个你想要迭代的集合；只需找到它并对其进行迭代。

于 2013-06-02T05:06:33.960 回答