我正在尝试使用请求上传文件.我需要上传一个PDF文件,同时将一些其他数据发送到表单,如作者姓名.
我试过这个:
requests.get(url, files = {"file":open("file.txt"), "author" : "me" })
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但它不会将数据发送到表单.
我正在学习算法..所以,我带来了一些非常有趣的东西.
线性方程((a*n)+b)的渐近界是O(n^2)......a>0.
这与不太令人惊讶的相同...... a* n^2 + b* n + c
为什么?
我有一个包含7000行字符串的文本文件.我必须根据少数参数搜索特定的字符串.
有人说下面的代码效率不高(速度和内存使用率).
f = open("file.txt")
data = f.read().split() # strings as list
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有点困惑
我有一个大的JSON对象(来自Google Feed API)和来自上下文的URL.当上下文中的URL与供稿条目URL中的URL相同时,django应该解析!
{for aEntry in feed.entries %}
{% if aEntry.link == {{my_URL}} %}
<p>URL FOUND in feed entry</p>
{% endif %}
{% endfor %}
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这不行!!
编辑:
feed: {
"feedUrl": "http://.blogspot.in//feeds/posts/default",
"title": "",
"link": "http://.blogspot.com/",
"author": "",
"description": "",
"type": "atom10",
"entries": [{
"title": "Shades",
"link": "http://.blogspot.com/myurl.html",
"author": "",
"publishedDate": "Sun, 29 Jul 2012 04:07:00 -0700",
"contentSnippet": "abstract art!",
"content": " HTML CONTENT HERE",
"categories": ["abstract"]
}
context = { "feed" : feed, "my_url" : "http://.blogspot.com/myurl.html"}
Run Code Online (Sandbox Code Playgroud) 假设我在Github中有以下结构(远程)
/project
/dir1
/dir2
file1
file2
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现在,我将repo分叉并将其更改为以下内容:
/project
/dir_list
/files_list
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当我推动回购时,我希望遥控器像:
/project
/dir_list
/files_list
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而不是:
/project
/dir1
/dir2
file1
file2
/dir_list
/files_list
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我怎么做?
# given an array, it solves sum of elements in it,
# args: array, initial and final indices
# Returns: sum of elements
def ArrayTot (arr, low, high):
return sum( arr[low : high+1] )
# Is this linear time?
# args: array, initial and final indices
# returns: Max possible value in sub-array.
def MaxSubArray (arr, low, high):
# max value of sub-array
TotalArray = 0
# starts iterating arr from left - right i.e., arr[low, j]
for j in …Run Code Online (Sandbox Code Playgroud)