Question

我想删除加倍的时间值（包括为其分配的测量值），并在两者之间填充缺失的时间值。

我正在处理一些模拟数据。基本上，在下面显示的示例中有列表。应删除倍增的时间元素，并为其分配指定的测量值。然后，应添加缺少的时间元素。缺少的测量值应由“-”代替。以后我可能会插值。

我尝试了一些事情，只是遍历列表，然后追加和插入，但这似乎不起作用。我也找不到解决方案的类似问题。

谢谢！

time = [1, 2, 3, 3, 5, 7, 7, 10]
meas = [10,20,30,40,50,60,70,80]

预期列表如下：

time = [1,2,3,4,5,6,7,8,9,10]
meas = [10 20 30 - 50 - 60 - - 80]

Answer 1

起草一个粗略的解决方案，可能会有战利品。但这有效。

time = [1, 2, 3, 3, 5, 7, 7, 10]
meas = [10,20,30,40,50,60,70,80]

# Remove duplicate time and sort result
t = list(set(time))
t.sort()

# Remove duplicate objects (keep 1st instance)
for n, e in enumerate(t):
    while e != time[n]:
        del time[n]
        del meas[n]

# Check that it works
print(time)
print(meas)

# Create list in the wanted format
m = 0  # keep track of meas index
meas_padded = []  # new list for result, could be done using the meas list but im to lazy right now.
for n in range(0, time[-1]):
    if n + 1 == time[m]:
        meas_padded.append(meas[m])
        m += 1
    else:
        meas_padded.append('--')

# As the list is padded to time[-1] size, I will not create the time series as it will map towars meas_padded index n+1
# Added padding and string repr since it was neat for debug.
print([str(n+1).zfill(2) for n, m in enumerate(meas_padded)])
print([str(m) for  m in meas_padded])

输出：

[1, 2, 3, 5, 7, 10]
[10, 20, 30, 50, 60, 80]
['01', '02', '03', '04', '05', '06', '07', '08', '09', '10']
['10', '20', '30', '--', '50', '--', '60', '--', '--', '80']

希望它可以帮助您：）

填充meas而不需要创建一个新的，可以使用带有reverse_range的一些技巧

Answer 2

您可以使用numpy来加快脚本速度：

import numpy as np

def get_correction(time, meas):
    """
    Deleting doubled measuring points and adding missing one

    :param np.array time: 1D array which contains times
    :param np.array time: 1D array with measurements for corresponding times
    :return: a tuple of corrected arrays (`time`,`meas`)
    """
    if type(t) == list: time = np.array(time)
    if type(m) == list: meas = np.array(meas)
    t_, index = np.unique(t, return_index=True) # Determine unique times and their indexes
    measures = np.zeros(np.max(t)) ; measures[:] = np.nan # Create empty measures array
    measures[t_ - 1] = m[index] # Insert correspondant values in measures array
    time = np.arange(np.max(t)) + 1 # Create correpondant times
    return (time, measures)

如果您测试此功能：

time = [1, 2, 3, 3, 5, 7, 7, 10]
meas = [10,20,30,40,50,60,70,80]

get_correction(time, meas)

您获得：

(array([ 1,  2,  3,  4,  5,  6,  7,  8,  9, 10]),
 array([10., 20., 30., nan, 50., nan, 60., nan, nan, 80.]))

删除加倍的测量点并添加缺失的一个

2 个答案: