Python：在列表

Question

考虑一个零和一的列表，其中一个代表“事件”：

signal = [0,0,0,0, 1,1, 0,0,0, 1,1,1, 0,0,0, 1, 0,0, 1,1,1,1, 0]

这里我们有4个不同持续时间的事件。什么是获得开始的最pythonic和整洁的方式（即第一个“1”的索引）和持续时间（即的数量）这些事件中的每一个？

我试过迭代列表，但我找不到找到持续时间的方法吗？

Answer 1

您可以使用itertools.groupby将连续事件分组在一起并计算1。它内存效率高，速度快。

from itertools import groupby

signals = [0,0,0,0,1,1,0,0,0,1,1,1,0,0,0,1,0,0,1,1,1,1,0]

def count_event(signals,event_value=1):
    """Counts no of events and counts the duration of the events

    Args:
        signal : list of signals
        event_value : value of the siganl

    Returns:
        list of tuples of index and duration of events
    """
    event_duration = []
    index = 0
    for key,g in (groupby(signal)):
        length = len(list(g))
        if key == event_value:
            event_duration.append((index,length))
        index += length
    return event_duration

print(count_event(signal,1))

输出：

[(4, 2), (9, 3), (15, 1), (18, 4)]

Answer 2

您可以使用itertools.groupby。我将onset，duration对放在元组列表中，但您可以使用以后会方便的任何数据结构：

>>> import itertools
>>> from operator import itemgetter
>>> gb = itertools.groupby(enumerate(signal), itemgetter(1))
>>> signals = []
>>> for k, g in gb:
...     if k:
...         sig = list(g)
...         onset = sig[0][0]
...         duration = len(sig)
...         signals.append((onset, duration))
...
>>> signals
[(4, 2), (9, 3), (15, 1), (18, 4)]

Answer 3

from pylab import *
x = array([0,0,0,0,1,1,0,0,0,1,1,1,0,0,0,1,0,0,1,1,1,1,0])
plus = where(x[1:] - x[:-1] > 0)[0]
minus = where(x[1:] - x[:-1] < 0)[0]
print vstack((plus + 1, minus -plus)).T