我在Python中使用BeautifulSoup来解析一些HTML。我正在处理的一个问题是，我遇到了colspans在标题行之间不同的情况。 （标题行是需要组合以在我的行话中获取列标题的行）这是一列可以跨越其上方或下方的多个列，并且需要根据跨越来追加或预先添加单词。以下是执行此操作的例程。我使用BeautifulSoup来拉动colspans并拉出每一行中每个单元格的内容。 longHeader是具有最多项目的标题行的内容，spanLong是包含行中每个项目的colspans的列表。这有效，但看起来并不像Pythonic。
Alos - 如果差异<0，它就不会起作用，我可以用我用来使其工作的相同方法来解决这个问题。但在此之前，我想知道是否有人能够快速看到这一点，并提出更多的Pythonic方法。我是一名长期以来的SAS程序员，因此我很难打破模具 - 我会编写代码，好像我正在编写SAS宏。
longHeader=['','','bananas','','','','','','','','','','trains','','planes','','','','']
shortHeader=['','','bunches','','cars','','trucks','','freight','','cargo','','all other','','']
spanShort=[1,1,3,1,3,1,3,1,3,1,3,1,3,1,3]
spanLong=[1,1,3,1,1,1,1,1,1,1,1,1,3,1,3,1,3,1,3]
combinedHeader=[]
sumSpanLong=0
sumSpanShort=0
spanDiff=0
longHeaderCount=0
for each in range(len(shortHeader)):
sumSpanLong=sumSpanLong+spanLong[longHeaderCount]
sumSpanShort=sumSpanShort+spanShort[each]
spanDiff=sumSpanShort-sumSpanLong
if spanDiff==0:
combinedHeader.append([longHeader[longHeaderCount]+' '+shortHeader[each]])
longHeaderCount=longHeaderCount+1
continue
for i in range(0,spanDiff):
combinedHeader.append([longHeader[longHeaderCount]+' '+shortHeader[each]])
longHeaderCount=longHeaderCount+1
sumSpanLong=sumSpanLong+spanLong[longHeaderCount]
spanDiff=sumSpanShort-sumSpanLong
if spanDiff==0:
combinedHeader.append([longHeader[longHeaderCount]+' '+shortHeader[each]])
longHeaderCount=longHeaderCount+1
break
print combinedHeader
答案 0 :(得分：3)
以下是您算法的修改版本。 zip 用于迭代短长度和标题，类对象用于计算和迭代长项目，以及组合标题。 而更适合内循环。 （原谅太短的名字）。
class collector(object):
def __init__(self, header):
self.longHeader = header
self.combinedHeader = []
self.longHeaderCount = 0
def combine(self, shortValue):
self.combinedHeader.append(
[self.longHeader[self.longHeaderCount]+' '+shortValue] )
self.longHeaderCount += 1
return self.longHeaderCount
def main():
longHeader = [
'','','bananas','','','','','','','','','','trains','','planes','','','','']
shortHeader = [
'','','bunches','','cars','','trucks','','freight','','cargo','','all other','','']
spanShort=[1,1,3,1,3,1,3,1,3,1,3,1,3,1,3]
spanLong=[1,1,3,1,1,1,1,1,1,1,1,1,3,1,3,1,3,1,3]
sumSpanLong=0
sumSpanShort=0
combiner = collector(longHeader)
for sLen,sHead in zip(spanShort,shortHeader):
sumSpanLong += spanLong[combiner.longHeaderCount]
sumSpanShort += sLen
while sumSpanShort - sumSpanLong > 0:
combiner.combine(sHead)
sumSpanLong += spanLong[combiner.longHeaderCount]
combiner.combine(sHead)
return combiner.combinedHeader
答案 1 :(得分：2)
在这个例子中，你真的有很多事情要做。
您已经“过度处理”了美丽的汤标记对象以制作列表。将它们保留为标签。
所有这些合并算法都很难。它有助于对待两个对称合并的东西。
这是一个应该直接使用Beautiful Soup Tag对象的版本。此外，此版本不会假设两行的长度。
def merge3( row1, row2 ):
i1= 0
i2= 0
result= []
while i1 != len(row1) or i2 != len(row2):
if i1 == len(row1):
result.append( ' '.join(row1[i1].contents) )
i2 += 1
elif i2 == len(row2):
result.append( ' '.join(row2[i2].contents) )
i1 += 1
else:
if row1[i1]['colspan'] < row2[i2]['colspan']:
# Fill extra cols from row1
c1= row1[i1]['colspan']
while c1 != row2[i2]['colspan']:
result.append( ' '.join(row2[i2].contents) )
c1 += 1
elif row1[i1]['colspan'] > row2[i2]['colspan']:
# Fill extra cols from row2
c2= row2[i2]['colspan']
while row1[i1]['colspan'] != c2:
result.append( ' '.join(row1[i1].contents) )
c2 += 1
else:
assert row1[i1]['colspan'] == row2[i2]['colspan']
pass
txt1= ' '.join(row1[i1].contents)
txt2= ' '.join(row2[i2].contents)
result.append( txt1 + " " + txt2 )
i1 += 1
i2 += 1
return result
答案 2 :(得分：1)
也许请查看部分问题的zip函数：
>>> execfile('so_ques.py')
[[' '], [' '], ['bananas bunches'], [' '], [' cars'], [' cars'], [' cars'], [' '], [' trucks'], [' trucks'], [' trucks'], [' '], ['trains freight'], [' '], ['planes cargo'], [' '], [' all other'], [' '], [' ']]
>>> zip(long_header, short_header)
[('', ''), ('', ''), ('bananas', 'bunches'), ('', ''), ('', 'cars'), ('', ''), ('', 'trucks'), ('', ''), ('', 'freight'), ('', ''), ('', 'cargo'), ('', ''), ('trains', 'all other'), ('', ''), ('planes', '')]
>>>
enumerate可以帮助避免使用计数器进行一些复杂的索引：
>>> diff_list = []
>>> for place, header in enumerate(short_header):
diff_list.append(abs(span_short[place] - span_long[place]))
>>> for place, num in enumerate(diff_list):
if num:
new_shortlist.extend(short_header[place] for item in range(num+1))
else:
new_shortlist.append(short_header[place])
>>> new_shortlist
['', '', 'bunches', '', 'cars', 'cars', 'cars', '', 'trucks', 'trucks', 'trucks', '',...
>>> z = zip(new_shortlist, long_header)
>>> z
[('', ''), ('', ''), ('bunches', 'bananas'), ('', ''), ('cars', ''), ('cars', ''), ('cars', '')...
更多的pythonic命名可以增加清晰度：
for each in range(len(short_header)):
sum_span_long += span_long[long_header_count]
sum_span_short += span_short[each]
span_diff = sum_span_short - sum_span_long
if not span_diff:
combined_header.append...
答案 3 :(得分：0)
我想我会回答我自己的问题，但我确实得到了很多帮助。感谢您的帮助。经过几次小修改后，我让S.LOTT回答了问题。 （它们可能很小，以至于不可见（内部笑话））。所以现在问题是为什么这更像Pythonic？我想我看到它不那么密集/使用原始输入而不是推导/我无法判断它是否更容易阅读---＆gt;虽然很容易阅读
row1=headerCells[0]
row2=headerCells[1]
i1= 0
i2= 0
result= []
while i1 != len(row1) or i2 != len(row2):
if i1 == len(row1):
result.append( ' '.join(row1[i1]) )
i2 += 1
elif i2 == len(row2):
result.append( ' '.join(row2[i2]) )
i1 += 1
else:
if int(row1[i1].get("colspan","1")) < int(row2[i2].get("colspan","1")):
c1= int(row1[i1].get("colspan","1"))
while c1 != int(row2[i2].get("colspan","1")):
txt1= ' '.join(row1[i1]) # needed to add when working adjust opposing case
txt2= ' '.join(row2[i2]) # needed to add when working adjust opposing case
result.append( txt1 + " " + txt2 ) # needed to add when working adjust opposing case
print 'stayed in middle', 'i1=',i1,'i2=',i2, ' c1=',c1
c1 += 1
i1 += 1 # Is this the problem it
elif int(row1[i1].get("colspan","1"))> int(row2[i2].get("colspan","1")):
# Fill extra cols from row2 Make same adjustment as above
c2= int(row2[i2].get("colspan","1"))
while int(row1[i1].get("colspan","1")) != c2:
result.append( ' '.join(row1[i1]) )
c2 += 1
i2 += 1
else:
assert int(row1[i1].get("colspan","1")) == int(row2[i2].get("colspan","1"))
pass
txt1= ' '.join(row1[i1])
txt2= ' '.join(row2[i2])
result.append( txt1 + " " + txt2 )
print 'went to bottom', 'i1=',i1,'i2=',i2
i1 += 1
i2 += 1
print result
答案 4 :(得分：0)
我现在有一个答案。我正在考虑这一点，并决定我需要使用每个答案的部分内容。我仍然需要弄清楚我是否想要一个类或一个函数。但我认为算法可能比任何其他算法都更像Pythonic。但是，它从一些非常慷慨的人提供的答案中大量借鉴。我很欣赏这些，因为我学到了很多东西。
为了节省必须制作测试用例的时间，我将在IDLE中粘贴我一直在使用的完整代码，然后使用HTML示例文件。除了做出关于类/函数的决定（我需要考虑如何在我的程序中使用此代码），我很高兴看到任何改进使代码更加Pythonic。
from BeautifulSoup import BeautifulSoup
original=file(r"C:\testheaders.htm").read()
soupOriginal=BeautifulSoup(original)
all_Rows=soupOriginal.findAll('tr')
header_Rows=[]
for each in range(len(all_Rows)):
header_Rows.append(all_Rows[each])
header_Cells=[]
for each in header_Rows:
header_Cells.append(each.findAll('td'))
temp_Header_Row=[]
header=[]
for row in range(len(header_Cells)):
for column in range(len(header_Cells[row])):
x=int(header_Cells[row][column].get("colspan","1"))
if x==1:
temp_Header_Row.append( ' '.join(header_Cells[row][column]) )
else:
for item in range(x):
temp_Header_Row.append( ''.join(header_Cells[row][column]) )
header.append(temp_Header_Row)
temp_Header_Row=[]
combined_Header=zip(*header)
for each in combined_Header:
print each
好的测试文件内容在下面对不起我试图附上这些但是无法实现：
<TABLE style="font-size: 10pt" cellspacing="0" border="0" cellpadding="0" width="100%">
<TR valign="bottom">
<TD width="40%"> </TD>
<TD width="5%"> </TD>
<TD width="3%"> </TD>
<TD width="3%"> </TD>
<TD width="1%"> </TD>
<TD width="5%"> </TD>
<TD width="3%"> </TD>
<TD width="3%"> </TD>
<TD width="1%"> </TD>
<TD width="5%"> </TD>
<TD width="3%"> </TD>
<TD width="1%"> </TD>
<TD width="1%"> </TD>
<TD width="5%"> </TD>
<TD width="3%"> </TD>
<TD width="1%"> </TD>
<TD width="1%"> </TD>
<TD width="5%"> </TD>
<TD width="3%"> </TD>
<TD width="3%"> </TD>
<TD width="1%"> </TD>
</TR>
<TR style="font-size: 10pt" valign="bottom">
<TD> </TD>
<TD> </TD>
<TD> </TD>
<TD> </TD>
<TD> </TD>
<TD> </TD>
<TD> </TD>
<TD> </TD>
<TD> </TD>
<TD> </TD>
<TD nowrap align="right" colspan="2">FOODS WE LIKE</TD>
<TD> </TD>
<TD> </TD>
<TD nowrap align="right" colspan="2"> </TD>
<TD> </TD>
<TD> </TD>
<TD nowrap align="right" colspan="2"> </TD>
<TD> </TD>
</TR>
<TR style="font-size: 10pt" valign="bottom">
<TD> </TD>
<TD> </TD>
<TD nowrap align="CENTER" colspan="6">SILLY STUFF</TD>
<TD> </TD>
<TD> </TD>
<TD nowrap align="right" colspan="2">OTHER THAN</TD>
<TD> </TD>
<TD> </TD>
<TD nowrap align="CENTER" colspan="6">FAVORITE PEOPLE</TD>
<TD> </TD>
</TR>
<TR style="font-size: 10pt" valign="bottom">
<TD> </TD>
<TD> </TD>
<TD nowrap align="right" colspan="2">MONTY PYTHON</TD>
<TD> </TD>
<TD> </TD>
<TD nowrap align="right" colspan="2">CHERRYPY</TD>
<TD> </TD>
<TD> </TD>
<TD nowrap align="right" colspan="2">APPLE PIE</TD>
<TD> </TD>
<TD> </TD>
<TD nowrap align="right" colspan="2">MOTHERS</TD>
<TD> </TD>
<TD> </TD>
<TD nowrap align="right" colspan="2">FATHERS</TD>
<TD> </TD>
</TR>
<TR style="font-size: 10pt" valign="bottom">
<TD nowrap align="left">Name</TD>
<TD> </TD>
<TD nowrap align="right" colspan="2">SHOWS</TD>
<TD> </TD>
<TD> </TD>
<TD nowrap align="right" colspan="2">PROGRAMS</TD>
<TD> </TD>
<TD> </TD>
<TD nowrap align="right" colspan="2">BANANAS</TD>
<TD> </TD>
<TD> </TD>
<TD nowrap align="right" colspan="2">PERFUME</TD>
<TD> </TD>
<TD> </TD>
<TD nowrap align="right" colspan="2">TOOLS</TD>
<TD> </TD>
</TR>
</TABLE>