python之如何用python去除元组列表

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。

请您参考如下方法:

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]