numpy(.ma) 数组：自上次值更改以来的值数量？

Question

我在 numpy.ma 数组中有一个状态信号（测量的热泵）以及时间戳。我想要的是它打开的时间长度和关闭的时间长度。 （不是每日运行时间或其他什么，这很容易..）

我拥有什么（示例，事实上我有超过 16 个月的分钟值..）：

Time    Value
12:00   0
12:01   1
12:02   1
...
12:29   1
12:30   1
12:31   0
...
12:41   0
12:42   1
...
13:01   1
13:02   0
...and so on

我想要什么作为输出：

running_time_was:
time   value (=minutes)
12:31  30
13:10  20

off_time_was:
time   value (=minutes)
12:42  11

(the exact timestamps doesn't matter so much, but there should be some)

我已经问过我的人知道谁了解Python（他们也不知道），并尝试在互联网上搜索，但我只是不知道要搜索什么。所以也许有人至少可以给我一个提示，我可以在谷歌中输入哪些单词？ :)

ps: 哇，stackoverflow 太棒了！作为被动用户，我已经对它的可用性感到惊讶了，但询问界面甚至更好:)

Answer 1

好吧，终于我得到了我自己。但非常感谢 Joe Kington 的回答，它极大地启发了我并教会了我所有的东西:)

脚本

import numpy as np

def main():
    # Generate some data...
    t = np.linspace(0, 10*np.pi, 30)
    x = np.sin(t)
    condition = np.where(x>0,1,0)
    
    onarray,offarray = on_off_times(condition)

    print "Condition: ",condition
    print "Ontimes:   ",onarray
    print "Offtimes:  ",offarray
    
        
def on_off_times(condition):
        
    changing=np.diff(condition)         #complete array, -1 when tunring off, +1 when turning on
    idx, = changing.nonzero()           #Indices of changepoints
    times=np.diff(idx)              #'Times' between changes
    times=np.r_[0,times]            # The first time can't be calculated ->is set to 0
    
    ontimes=np.where(changing[idx]<0,times,False)   #When turning off: Was Ontime (times-array with "False" instead of offtimes)
    offtimes=np.where(changing[idx]>0,times,False)  #When turning on:  Was Offtime
    
    onarray=np.r_[changing.copy(),0]            #copy the array with all the values and add 1(to have an "empty" array of the right size)
    offarray=np.r_[changing.copy(),0]
    
    np.put(onarray,idx,ontimes)         #Put the times into it (at the changepoints)
    np.put(offarray,idx,offtimes)

    return onarray,offarray
    
main()

产量：

Condition:  [0 1 1 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0]
Ontimes:    [0 0 2 0 0 0 0 0 3 0 0 0 0 0 3 0 0 0 0 0 3 0 0 0 0 0 3 0 0 0]
Offtimes:   [0 0 0 0 0 3 0 0 0 0 0 3 0 0 0 0 0 3 0 0 0 0 0 3 0 0 0 0 0 0]

值得一提的事情：

• 是的，代码按照我在问题中的要求提供了另一个输出..
  抱歉，不精确，但该函数给出的输出是什么
  我终于需要了。
• 现在我只需要得到输出中的零
  蒙面，这很容易。
• 如果有人有更好的方法将数据放入新数组中，请编辑:)
• 如何链接到用户 Joe？或者如何感谢他？

Answer 2

基本上，您有一个布尔数组，并且想要找到连续区域的起点和终点。

最好避免循环遍历 numpy 数组的每个项目。

有几种不同的方法可以做到这一点，但我通常会做类似的事情（我最初可能是从 这里)：

import numpy as np
def contiguous_regions(condition):
    """Finds contiguous True regions of the boolean array "condition". Returns
    a 2D array where the first column is the start index of the region and the
    second column is the end index."""

    # Find the indicies of changes in "condition"
    idx, = np.diff(condition).nonzero()

    # Prepend or append the start or end indicies to "idx"
    # if there's a block of "True"'s at the start or end...
    if condition[0]:
        idx = np.r_[0, idx]
    if condition[-1]:
        idx = np.r_[idx, len(condition)-1]

    return idx.reshape((-1,2))

作为一个简单的例子：

import numpy as np

def main():
    # Generate some data...
    t = np.linspace(0, 6*np.pi, 100)
    x = np.sin(t)
    condition = x > 0

    regions = contiguous_regions(condition)
    lengths = regions[:,1] - regions[:,0]

    for reg, length in zip(regions, lengths):
        print 'Condition was True for {0} seconds'.format(length)
        print '    From time {0}s to {1}s'.format(*reg)

def contiguous_regions(condition):
    idx, = np.diff(condition).nonzero()

    if condition[0]:
        idx = np.r_[0, idx]
    if condition[-1]:
        idx = np.r_[idx, len(condition)-1]

    return idx.reshape((-1,2))

main()

这会产生：

Condition was True for 16 seconds
    From time 0s to 16s
Condition was True for 16 seconds
    From time 33s to 49s
Condition was True for 16 seconds
    From time 66s to 82s

Answer 3

我会按照以下方式完成此操作：

# read data from file
dt = {'names' : ('ts', 'state'), 'formats' : ('S5','i4')}
data = np.loadtxt(datafn, dtype = dt)
# minutes counter
mc = 1
# current state, current timestamp
cs = data[0]['state']
ct = data[0]['ts']
# states dictionary
states = {1 : [], 0 : []}

for minute in data[1:]:
    if cs != minute['state']:
        states[cs].append([ct, mc])
        mc = 0
        cs = minute['state']
        ct = minute['ts']
    else:
        mc += 1
# Printing the result
print 'On time'
for [ts, mc] in states[1]:
     print '%s\t%i' % (ts, mc)
print 'Off time'
for [ts, mc] in states[0]:
     print '%s\t%i' % (ts, mc)

极其未经测试，但您可以了解逻辑。

Answer 4

（可能）答案

您可以尝试以下操作：

Since = 0
for i in range(1, Data.shape[0]):
    #Switched off
    if Data[i, 1] == 0.0 and Data[i - 1, 1] == 1.0:
        print "{0} for {1}min".format(Data[i, 0], i - Since)
    #Switched on
    elif Data[i, 1] == 1.0 and Data[i - 1, 1] == 0.0:
        Since = i

您循环遍历空数组 (Data)，该数组的第一列中包含时间戳，第二列中包含 1.0 或 < code>0.0 取决于加热器是否打开或关闭。

您可以通过查看实际的开/关值和前一个值来检测状态的变化。根据这两个值，您可以看到加热器是关闭还是开启。然后您需要做的就是将当前索引的值保存在 Since 中，然后您就可以获得加热器打开的时间。

脚本

使用以下脚本，您可以设置一个数据数组并运行上面的代码，看看它是如何工作的：

import datetime
import numpy as np

#Setting up OnOff array
OnOff = np.concatenate((np.zeros((7,)), np.ones((20,)), np.zeros((3,)), np.ones((5,)), np.zeros((4,)), np.ones((16,)), np.zeros((2,)), np.ones((2,)), np.zeros((1,))))

#Setting up time array
start = datetime.time(12, 00)
TimeStamps = []

for i in range(OnOff.size):
    TimeStamps.append(datetime.time(12 + i/60, np.mod(i, 60)))

TimeStamps = np.array(TimeStamps)

#Concatenating both arrays to a single array
Data = np.hstack((np.reshape(TimeStamps, (TimeStamps.size, 1)), np.reshape(OnOff, (OnOff.size, 1))))

Since = 0
for i in range(1, Data.shape[0]):
    #Switched off
    if Data[i, 1] == 0.0 and Data[i - 1, 1] == 1.0:
        print "{0} for {1}min".format(Data[i, 0], i - Since)
    #Switched on
    elif Data[i, 1] == 1.0 and Data[i - 1, 1] == 0.0:
        Since = i

输出是

12:27:00 for 20min
12:35:00 for 5min
12:55:00 for 16min
12:59:00 for 2min