Python计算列表中的分组项目

Question

问题的例子。

我有一个列表，其中某些元素包含粗体标签，这些标签标识其他元素的分组。 例如：

list_a = ['1','<b>a</b>','2','3','<b>bi</b>','<b>bii</b>','4', '<b>c</b>', '5']

我想知道非粗体字符串如何在粗体​​标签之间分组。 我的输出将是：

grouped = [2,1,1]

编辑：我的实际代码/问题如下。

我的代码。

import scraperwiki
import urllib2
import re

# URL for scraping - Use sparingly.
url = 'http://www.cmc.gv.ao/sites/main/pt/Lists/CMC%20%20PublicaesFicheiros/Attachments/89/Lista%20de%20Institui%C3%A7%C3%B5es%20Registadas%20(actualizado%2004.07.16).pdf'

# Open URL, read information.
u = urllib2.urlopen(url)
xml=scraperwiki.pdftoxml(u.read()) # interpret as xml
u.close()

# Get text from xml.
text = re.findall(r'>(.*)</text>', xml)

# Remove empty strings and empty tags.  
text = [i.strip() for i in text if not len(i) == 1 if not i == '<b> </b>']

# Replace characters as needed.
text = [i.replace('&amp;','&') for i in text]

# Remove title and headers.
start = 0
for i in text:
    if '<b>' not in i: start += 1
    else: break

# This is the list.
text = text[start:]

我当前的工作解决方案。

# Work out how the records are grouped together.
count = 0
grouped = []
for i in text:
    # If count is zero no grouping has yet been counted - nothing to count.
    if '<b>' in i and count != 0:
        # Append the number showing how many records are in the group.
        grouped.append(count)
        # Reset the counter before counting items in the next group.
        count = 0

    elif i.isnumeric():
        count += 1

    else: pass

# Append the count of the final grouping.
grouped.append(count)

有没有更好/更多的pythonic方式来做到这一点？

Answer 1

您可以在此处使用itertools ：

from itertools import dropwhile, groupby

list_a = ['1','<b>a</b>','2','3','<b>bi</b>','<b>bii</b>','4', '<b>c</b>', '5']
res = [len(list(g)) for k, g in groupby(dropwhile(lambda L: '<b>' not in L, list_a), str.isnumeric) if k]
# [2, 1, 1]

分解以使事情更清楚：

from itertools import dropwhile, groupby

def count(iterable):
    # Ignore everything until the first `<b>`
    start_at = dropwhile(lambda L: '<b>' in L, iterable)
    # Groupby whether it's a numeric or not 
    grouped = groupby(start_at, str.isnumeric)
    # Loop over each key and group (key will be True for numeric values)
    for k, g in grouped:
        # The group will contain one or more numeric values
        if k:
            # yield the length of the group
            yield len(list(g))

res = list(count(list_a))

Answer 2

from itertools import *

def iter_len(i):
    return sum(1 for _ in i)

print [
    iter_len(v)
    for k, v in groupby(text, key=lambda x: '<b>' in x)
    if not k
]