Create List of Dictionary Python


Question

I want to get all the iframe from a webpage.

Code:

site = "http://" + url
f = urllib2.urlopen(site)
web_content =  f.read()

soup = BeautifulSoup(web_content)
info = {}
content = []
for iframe in soup.find_all('iframe'):
    info['src'] = iframe.get('src')
    info['height'] = iframe.get('height')
    info['width'] = iframe.get('width')
    content.append(info)
    print(info)       

pprint(content)

result of print(info):

{'src': u'abc.com', 'width': u'0', 'height': u'0'}
{'src': u'xyz.com', 'width': u'0', 'height': u'0'}
{'src': u'http://www.detik.com', 'width': u'1000', 'height': u'600'}

result of pprint(content):

[{'height': u'600', 'src': u'http://www.detik.com', 'width': u'1000'},
{'height': u'600', 'src': u'http://www.detik.com', 'width': u'1000'},
{'height': u'600', 'src': u'http://www.detik.com', 'width': u'1000'}]

Why is the value of the content not right? It's suppose to be the same as the value when I print(info).

1
27
11/18/2016 2:25:49 PM

Accepted Answer

You are not creating a separate dictionary for each iframe, you just keep modifying the same dictionary over and over, and you keep adding additional references to that dictionary in your list.

Remember, when you do something like content.append(info), you aren't making a copy of the data, you are simply appending a reference to the data.

You need to create a new dictionary for each iframe.

for iframe in soup.find_all('iframe'):
   info = {}
    ...

Even better, you don't need to create an empty dictionary first. Just create it all at once:

for iframe in soup.find_all('iframe'):
    info = {
        "src":    iframe.get('src'),
        "height": iframe.get('height'),
        "width":  iframe.get('width'),
    }
    content.append(info)

There are other ways to accomplish this, such as iterating over a list of attributes, or using list or dictionary comprehensions, but it's hard to improve upon the clarity of the above code.

54
7/15/2012 2:37:56 PM

You have misunderstood the Python list object. It is similar to a C pointer-array. It does not actually "copy" the object which you append to it. Instead, it just store a "pointer" to that object.

Try the following code:

>>> d={}
>>> dlist=[]
>>> for i in xrange(0,3):
    d['data']=i
    dlist.append(d)
    print(d)

{'data': 0}
{'data': 1}
{'data': 2}
>>> print(dlist)
[{'data': 2}, {'data': 2}, {'data': 2}]

So why is print(dlist) not the same as print(d)?

The following code shows you the reason:

>>> for i in dlist:
    print "the list item point to object:", id(i)

the list item point to object: 47472232
the list item point to object: 47472232
the list item point to object: 47472232

So you can see all the items in the dlist is actually pointing to the same dict object.

The real answer to this question will be to append the "copy" of the target item, by using d.copy().

>>> dlist=[]
>>> for i in xrange(0,3):
    d['data']=i
    dlist.append(d.copy())
    print(d)

{'data': 0}
{'data': 1}
{'data': 2}
>>> print dlist
[{'data': 0}, {'data': 1}, {'data': 2}]

Try the id() trick, you can see the list items actually point to completely different objects.

>>> for i in dlist:
    print "the list item points to object:", id(i)

the list item points to object: 33861576
the list item points to object: 47472520
the list item points to object: 47458120

Licensed under: CC-BY-SA with attribution
Not affiliated with: Stack Overflow
Icon