BeautifulSoup tag.children only gets odd-numbered elements
I want to move the elements from one tag to another tag using the following code:
soup = BeautifulSoup("<p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p>")
d = soup.new_tag('div')
for tag in soup.body.children:
d.append(tag)
However, this yields...
>>> d
<div><p>I wish I was bold.</p><p>me three</p><p>5</p></div>
Only the odd-numbered elements were moved. I checked soup.body.children to see what it looks like (before moving anything), and all of the tags appear to be there:
>>> list(soup.body.children)
[<p>I wish I was bold.</p>, <p>me too</p>, <p>me three</p>, <p>me 4</p>, <p>5</p>]
When I iterate over list(soup.body.children), then everything works as expected:
soup = BeautifulSoup("<p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p>")
d = soup.new_tag('div')
for tag in list(soup.body.children):
d.append(tag)
>>> d
<div><p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p></div>
Why does iterating over soup.body.children only access odd-numbered tags, but iterating over list(soup.body.children) accesses all of them?
python beautifulsoup
asked Nov 20 '18 at 19:28
reynoldsnlpreynoldsnlp
6591826
add a comment |
I want to move the elements from one tag to another tag using the following code:
soup = BeautifulSoup("<p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p>")
d = soup.new_tag('div')
for tag in soup.body.children:
d.append(tag)
However, this yields...
>>> d
<div><p>I wish I was bold.</p><p>me three</p><p>5</p></div>
Only the odd-numbered elements were moved. I checked soup.body.children to see what it looks like (before moving anything), and all of the tags appear to be there:
>>> list(soup.body.children)
[<p>I wish I was bold.</p>, <p>me too</p>, <p>me three</p>, <p>me 4</p>, <p>5</p>]
When I iterate over list(soup.body.children), then everything works as expected:
soup = BeautifulSoup("<p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p>")
d = soup.new_tag('div')
for tag in list(soup.body.children):
d.append(tag)
>>> d
<div><p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p></div>
Why does iterating over soup.body.children only access odd-numbered tags, but iterating over list(soup.body.children) accesses all of them?
python beautifulsoup
asked Nov 20 '18 at 19:28
reynoldsnlpreynoldsnlp
6591826
Which parser are you using?
– Stephen Cowley
Nov 20 '18 at 19:33
@StephenCowleylxml, but why would the parser make a difference? I thought that once you've parsed the markup, it's all just native soup objects.
– reynoldsnlp
Nov 20 '18 at 19:38
Yeah, I wasn't sure if it mattered, but just wanted to match your configuration as close as possible.
– Stephen Cowley
Nov 20 '18 at 19:44
add a comment |
I want to move the elements from one tag to another tag using the following code:
soup = BeautifulSoup("<p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p>")
d = soup.new_tag('div')
for tag in soup.body.children:
d.append(tag)
However, this yields...
>>> d
<div><p>I wish I was bold.</p><p>me three</p><p>5</p></div>
Only the odd-numbered elements were moved. I checked soup.body.children to see what it looks like (before moving anything), and all of the tags appear to be there:
>>> list(soup.body.children)
[<p>I wish I was bold.</p>, <p>me too</p>, <p>me three</p>, <p>me 4</p>, <p>5</p>]
When I iterate over list(soup.body.children), then everything works as expected:
soup = BeautifulSoup("<p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p>")
d = soup.new_tag('div')
for tag in list(soup.body.children):
d.append(tag)
>>> d
<div><p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p></div>
Why does iterating over soup.body.children only access odd-numbered tags, but iterating over list(soup.body.children) accesses all of them?
python beautifulsoup
asked Nov 20 '18 at 19:28
reynoldsnlpreynoldsnlp
6591826
I want to move the elements from one tag to another tag using the following code:
soup = BeautifulSoup("<p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p>")
d = soup.new_tag('div')
for tag in soup.body.children:
d.append(tag)
However, this yields...
>>> d
<div><p>I wish I was bold.</p><p>me three</p><p>5</p></div>
Only the odd-numbered elements were moved. I checked soup.body.children to see what it looks like (before moving anything), and all of the tags appear to be there:
>>> list(soup.body.children)
[<p>I wish I was bold.</p>, <p>me too</p>, <p>me three</p>, <p>me 4</p>, <p>5</p>]
When I iterate over list(soup.body.children), then everything works as expected:
soup = BeautifulSoup("<p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p>")
d = soup.new_tag('div')
for tag in list(soup.body.children):
d.append(tag)
>>> d
<div><p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p></div>
Why does iterating over soup.body.children only access odd-numbered tags, but iterating over list(soup.body.children) accesses all of them?
python beautifulsoup
python beautifulsoup
asked Nov 20 '18 at 19:28
reynoldsnlpreynoldsnlp
6591826
asked Nov 20 '18 at 19:28
reynoldsnlpreynoldsnlp
6591826
asked Nov 20 '18 at 19:28
reynoldsnlpreynoldsnlp
6591826
asked Nov 20 '18 at 19:28
reynoldsnlpreynoldsnlp
6591826
asked Nov 20 '18 at 19:28
reynoldsnlpreynoldsnlp
6591826
6591826
Which parser are you using?
– Stephen Cowley
Nov 20 '18 at 19:33
@StephenCowleylxml, but why would the parser make a difference? I thought that once you've parsed the markup, it's all just native soup objects.
– reynoldsnlp
Nov 20 '18 at 19:38
Yeah, I wasn't sure if it mattered, but just wanted to match your configuration as close as possible.
– Stephen Cowley
Nov 20 '18 at 19:44
add a comment |
Which parser are you using?
– Stephen Cowley
Nov 20 '18 at 19:33
@StephenCowleylxml, but why would the parser make a difference? I thought that once you've parsed the markup, it's all just native soup objects.
– reynoldsnlp
Nov 20 '18 at 19:38
Yeah, I wasn't sure if it mattered, but just wanted to match your configuration as close as possible.
– Stephen Cowley
Nov 20 '18 at 19:44
Which parser are you using?
– Stephen Cowley
Nov 20 '18 at 19:33
Which parser are you using?
– Stephen Cowley
Nov 20 '18 at 19:33
@StephenCowley
lxml, but why would the parser make a difference? I thought that once you've parsed the markup, it's all just native soup objects.– reynoldsnlp
Nov 20 '18 at 19:38
@StephenCowley
lxml, but why would the parser make a difference? I thought that once you've parsed the markup, it's all just native soup objects.– reynoldsnlp
Nov 20 '18 at 19:38
Yeah, I wasn't sure if it mattered, but just wanted to match your configuration as close as possible.
– Stephen Cowley
Nov 20 '18 at 19:44
Yeah, I wasn't sure if it mattered, but just wanted to match your configuration as close as possible.
– Stephen Cowley
Nov 20 '18 at 19:44
add a comment |
2 Answers
2
active
oldest
votes
When you append to the d tag in the first case, you are actually changing the size of soup.body.children as you go, since it moves the tags from soup to d.
Thus it grabs the tag at 0 and moves it to d. When it goes back for tag at 1, they have all shifted over, and it grabs the tag originally at index 2.
One way to see this in action is to actually print list(soup.body.children) in each iteration. Something like this:
for i, tag in enumerate(soup.body.children):
d.append(tag)
print(i)
print(list(soup.body.children))
print()
Output:
0 #<-- It's accessing this element
[<p>me too</p>, <p>me three</p>, <p>me 4</p>, <p>5</p>]
1
[<p>me too</p>, <p>me 4</p>, <p>5</p>]
2
[<p>me too</p>, <p>me 4</p>]
answered Nov 20 '18 at 19:38
Stephen CowleyStephen Cowley
1,164418
1
Of course! That's why making a copy withlistdoesn't fail. Well done.
– reynoldsnlp
Nov 20 '18 at 19:40
add a comment |
It seems like the items being moved from 'soup' to 'd' are affected the iterator, i.e. removing them and changing the 'soup' generator as you iterate through the items.
soup = BeautifulSoup("<p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p>")
d = soup.new_tag('div')
children = list(soup.body.children).copy()
for tag in children:
d.append(tag)
print(d)
Creating a copy of the list solves the problem.
output
<div><p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p></div>
answered Nov 20 '18 at 19:39
Victor 'Chris' CabralVictor 'Chris' Cabral
1,4991221
1
.copy()is unnecessary, sincelist()already makes a copy, as I showed in the OP.
– reynoldsnlp
Nov 20 '18 at 19:47
add a comment |
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2 Answers
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2 Answers
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oldest
votes
When you append to the d tag in the first case, you are actually changing the size of soup.body.children as you go, since it moves the tags from soup to d.
Thus it grabs the tag at 0 and moves it to d. When it goes back for tag at 1, they have all shifted over, and it grabs the tag originally at index 2.
One way to see this in action is to actually print list(soup.body.children) in each iteration. Something like this:
for i, tag in enumerate(soup.body.children):
d.append(tag)
print(i)
print(list(soup.body.children))
print()
Output:
0 #<-- It's accessing this element
[<p>me too</p>, <p>me three</p>, <p>me 4</p>, <p>5</p>]
1
[<p>me too</p>, <p>me 4</p>, <p>5</p>]
2
[<p>me too</p>, <p>me 4</p>]
answered Nov 20 '18 at 19:38
Stephen CowleyStephen Cowley
1,164418
1
Of course! That's why making a copy withlistdoesn't fail. Well done.
– reynoldsnlp
Nov 20 '18 at 19:40
add a comment |
When you append to the d tag in the first case, you are actually changing the size of soup.body.children as you go, since it moves the tags from soup to d.
Thus it grabs the tag at 0 and moves it to d. When it goes back for tag at 1, they have all shifted over, and it grabs the tag originally at index 2.
One way to see this in action is to actually print list(soup.body.children) in each iteration. Something like this:
for i, tag in enumerate(soup.body.children):
d.append(tag)
print(i)
print(list(soup.body.children))
print()
Output:
0 #<-- It's accessing this element
[<p>me too</p>, <p>me three</p>, <p>me 4</p>, <p>5</p>]
1
[<p>me too</p>, <p>me 4</p>, <p>5</p>]
2
[<p>me too</p>, <p>me 4</p>]
answered Nov 20 '18 at 19:38
Stephen CowleyStephen Cowley
1,164418
1
Of course! That's why making a copy withlistdoesn't fail. Well done.
– reynoldsnlp
Nov 20 '18 at 19:40
add a comment |
When you append to the d tag in the first case, you are actually changing the size of soup.body.children as you go, since it moves the tags from soup to d.
Thus it grabs the tag at 0 and moves it to d. When it goes back for tag at 1, they have all shifted over, and it grabs the tag originally at index 2.
One way to see this in action is to actually print list(soup.body.children) in each iteration. Something like this:
for i, tag in enumerate(soup.body.children):
d.append(tag)
print(i)
print(list(soup.body.children))
print()
Output:
0 #<-- It's accessing this element
[<p>me too</p>, <p>me three</p>, <p>me 4</p>, <p>5</p>]
1
[<p>me too</p>, <p>me 4</p>, <p>5</p>]
2
[<p>me too</p>, <p>me 4</p>]
answered Nov 20 '18 at 19:38
Stephen CowleyStephen Cowley
1,164418
When you append to the d tag in the first case, you are actually changing the size of soup.body.children as you go, since it moves the tags from soup to d.
Thus it grabs the tag at 0 and moves it to d. When it goes back for tag at 1, they have all shifted over, and it grabs the tag originally at index 2.
One way to see this in action is to actually print list(soup.body.children) in each iteration. Something like this:
for i, tag in enumerate(soup.body.children):
d.append(tag)
print(i)
print(list(soup.body.children))
print()
Output:
0 #<-- It's accessing this element
[<p>me too</p>, <p>me three</p>, <p>me 4</p>, <p>5</p>]
1
[<p>me too</p>, <p>me 4</p>, <p>5</p>]
2
[<p>me too</p>, <p>me 4</p>]
answered Nov 20 '18 at 19:38
Stephen CowleyStephen Cowley
1,164418
edited Nov 20 '18 at 19:48
answered Nov 20 '18 at 19:38
Stephen CowleyStephen Cowley
1,164418
answered Nov 20 '18 at 19:38
Stephen CowleyStephen Cowley
1,164418
answered Nov 20 '18 at 19:38
Stephen CowleyStephen Cowley
1,164418
1,164418
1
Of course! That's why making a copy withlistdoesn't fail. Well done.
– reynoldsnlp
Nov 20 '18 at 19:40
add a comment |
1
Of course! That's why making a copy withlistdoesn't fail. Well done.
– reynoldsnlp
Nov 20 '18 at 19:40
1
1
Of course! That's why making a copy with
list doesn't fail. Well done.– reynoldsnlp
Nov 20 '18 at 19:40
Of course! That's why making a copy with
list doesn't fail. Well done.– reynoldsnlp
Nov 20 '18 at 19:40
add a comment |
It seems like the items being moved from 'soup' to 'd' are affected the iterator, i.e. removing them and changing the 'soup' generator as you iterate through the items.
soup = BeautifulSoup("<p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p>")
d = soup.new_tag('div')
children = list(soup.body.children).copy()
for tag in children:
d.append(tag)
print(d)
Creating a copy of the list solves the problem.
output
<div><p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p></div>
answered Nov 20 '18 at 19:39
Victor 'Chris' CabralVictor 'Chris' Cabral
1,4991221
1
.copy()is unnecessary, sincelist()already makes a copy, as I showed in the OP.
– reynoldsnlp
Nov 20 '18 at 19:47
add a comment |
It seems like the items being moved from 'soup' to 'd' are affected the iterator, i.e. removing them and changing the 'soup' generator as you iterate through the items.
soup = BeautifulSoup("<p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p>")
d = soup.new_tag('div')
children = list(soup.body.children).copy()
for tag in children:
d.append(tag)
print(d)
Creating a copy of the list solves the problem.
output
<div><p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p></div>
answered Nov 20 '18 at 19:39
Victor 'Chris' CabralVictor 'Chris' Cabral
1,4991221
1
.copy()is unnecessary, sincelist()already makes a copy, as I showed in the OP.
– reynoldsnlp
Nov 20 '18 at 19:47
add a comment |
It seems like the items being moved from 'soup' to 'd' are affected the iterator, i.e. removing them and changing the 'soup' generator as you iterate through the items.
soup = BeautifulSoup("<p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p>")
d = soup.new_tag('div')
children = list(soup.body.children).copy()
for tag in children:
d.append(tag)
print(d)
Creating a copy of the list solves the problem.
output
<div><p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p></div>
answered Nov 20 '18 at 19:39
Victor 'Chris' CabralVictor 'Chris' Cabral
1,4991221
It seems like the items being moved from 'soup' to 'd' are affected the iterator, i.e. removing them and changing the 'soup' generator as you iterate through the items.
soup = BeautifulSoup("<p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p>")
d = soup.new_tag('div')
children = list(soup.body.children).copy()
for tag in children:
d.append(tag)
print(d)
Creating a copy of the list solves the problem.
output
<div><p>I wish I was bold.</p><p>me too</p><p>me three</p><p>me 4</p><p>5</p></div>
answered Nov 20 '18 at 19:39
Victor 'Chris' CabralVictor 'Chris' Cabral
1,4991221
answered Nov 20 '18 at 19:39
Victor 'Chris' CabralVictor 'Chris' Cabral
1,4991221
answered Nov 20 '18 at 19:39
Victor 'Chris' CabralVictor 'Chris' Cabral
1,4991221
answered Nov 20 '18 at 19:39
Victor 'Chris' CabralVictor 'Chris' Cabral
1,4991221
1,4991221
1
.copy()is unnecessary, sincelist()already makes a copy, as I showed in the OP.
– reynoldsnlp
Nov 20 '18 at 19:47
add a comment |
1
.copy()is unnecessary, sincelist()already makes a copy, as I showed in the OP.
– reynoldsnlp
Nov 20 '18 at 19:47
1
1
.copy() is unnecessary, since list() already makes a copy, as I showed in the OP.– reynoldsnlp
Nov 20 '18 at 19:47
.copy() is unnecessary, since list() already makes a copy, as I showed in the OP.– reynoldsnlp
Nov 20 '18 at 19:47
add a comment |
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Which parser are you using?
– Stephen Cowley
Nov 20 '18 at 19:33
@StephenCowley
lxml, but why would the parser make a difference? I thought that once you've parsed the markup, it's all just native soup objects.– reynoldsnlp
Nov 20 '18 at 19:38
Yeah, I wasn't sure if it mattered, but just wanted to match your configuration as close as possible.
– Stephen Cowley
Nov 20 '18 at 19:44