1

对于每种情况，我都有一个带有一些标志的数组。为了在 HTML 中使用打印数组并使用 colspan，我需要将其转换为：

[{'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': True, 'open': True}, {'serve': False, 'open': True}, {'serve': False, 'open': True}, {'serve': False, 'open': True}, {'serve': False, 'open': True}, {'serve': False, 'open': True}, {'serve': False, 'open': True}, {'serve': False, 'open': True}, {'serve': False, 'open': True}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}, {'serve': False, 'open': False}]

在此打开标志：

[{'colspan': 12, 'open': False}, {'colspan': 60, 'open': True}, {'colspan': 24, 'open': False}]

另一个生成发球。

如何使用 Python 以最聪明的方式做到这一点？

我可以一一数一数，但这似乎不是一个好主意。

4

2 回答 2

4 
def cluster(dicts, key):
    current_value = None
    current_span = 0
    result = []

    for d in dicts:
        value = d[key]
        if current_value is None:
            current_value = value
        elif current_value != value:
            result.append({'colspan': current_span, key: current_value})
            current_value = value
            current_span = 0
        current_span += 1

    result.append({'colspan': current_span, key: current_value})
    return result

by_open = cluster(data, 'open')
by_serve = cluster(data, 'serve')

第二个版本，灵感来自丹尼斯的回答和他对itertools.groupby:

import itertools
import operator

def make_spans(data, key):
    groups = itertools.groupby(data, operator.itemgetter(key))
    return [{'colspan': len(list(items)), key: value} for value, items in groups]
于 2009-10-12T08:06:54.667 回答
4

这不清楚您需要什么，但我希望以下示例对您有所帮助：

>>> groupped = itertools.groupby(your_list, operator.itemgetter('open'))
>>> [{'colspan': len(list(group)), 'open': open} for open, group in groupped]
[{'colspan': 12, 'open': False}, {'colspan': 60, 'open': True}, {'colspan': 78, 'open': False}]
>>> groupped = itertools.groupby(your_list)
>>> [dict(d, colspan=len(list(group))) for d, group in groupped]
[{'serve': False, 'open': False, 'colspan': 12}, {'serve': True, 'open': True, 'colspan': 52}, {'serve': False, 'open': True, 'colspan': 8}, {'serve': False, 'open': False, 'colspan': 78}]
于 2009-10-12T08:18:48.297 回答