python - 如何用python去除元组列表？

Question

我有一个数组，每个案例都有一些标志。 为了在 HTML 中使用打印数组并使用 colspan，我需要转换它:

[{'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}]

在此为打开标志:

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

另一个生成发球。

我怎样才能使用 Python 以最聪明的方式做到这一点？

我可以一个一个地数一数，但这并不是一个好主意。

Answer 1

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 Denis 的回答和他对 itertools.groupby 的使用:

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]