如何用python剥离元组列表？答案

【问题标题】：How to strip a list of tuple with python?如何用python剥离元组列表？
【发布时间】：2010-12-05 21:55:10
【问题描述】：

对于每种情况，我都有一个带有一些标志的数组。为了使用在 HTML 中打印数组并使用 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}]

在此为打开标志：

[{'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，但这似乎不是个好主意。

【问题讨论】：

您的问题还不清楚。此外，这些 colspan 值是 freaky
这个 colspan 值用于这个简单的例子。 colspan 值是具有相同标志的连续案例的数量。

标签： python list dictionary flags

【解决方案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]

【讨论】：

colspan 每次都应该重新启动。这个函数给我 [{'colspan': 12, 'open': False}, {'colspan': 72, 'open': True}, {'colspan': 96, 'open': False}]
我的代码中缺少一行，我在几分钟前更正了 (current_span = 0)。它现在应该可以按预期工作。请原谅我的问题，但你知道如何编程，对吧？我的意思是，代码只是让您入门的示例，而不是生产就绪代码。如果您了解代码在做什么，那么发现您描述的错误应该很容易！
是的，我自己也修好了 :)

【解决方案2】：

这个不清楚你需要什么，但希望下面的例子对你有所帮助：

>>> 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}]

【讨论】：

不知道为什么最后一个colspan没有预期值。
好吧，我知道……我的示例得到了 150 个案例而不是 96 个……当您知道如何使用 Python 时，它是如此强大。谢谢。
+1 包括电池！使用itertools.groupby 是要走的路！
如果我想保留一个我知道对所有组相同的信息，我该怎么办？例如：[{'colspan': 12, 'open': False}, {'colspan': 60, 'open': True, 'schedule_id':1}, {'colspan': 78, 'open': False }]