Question

#Find values that are in range
in_range = [lo_lim <= v <= hi_lim for v in values]
#Find runs of in-range values
runs = [sum(1 for _ in group) for v, group in groupby(in_range) if v]

#Estimate total time spent in-range
total_time = sum(v if v > 1 else (Buffer_Value*sample_rate) for v in runs)

我正在尝试增加此代码以获取两组值，以及2对hi / lo限制，计算在这些限制范围内花费的总时间，以及在限制情况下的组合限制＆＃39;在同一点上也是如此，即

如果有100个数据点（两个数据集的长度相同，请检查每个点，

if values_1[45] and values_2[45] are in their respective limits

然后按范围计算。基本上将此if转换为生成器表达式：

if lo_lim_1<=Data_Points_1[i]<=hi_lim_1 and lo_lim_2<=Data_Points_2[i]<=hi_lim_2:

计算运行次数，如果运行长度是一个数据点，则应用缓冲区，否则应用采样率转换。

Answer 1

如果我理解你的问题，这应该有效。基本思路是将zip两个序列组合成对应的值对，然后使用and操作查找它们都在相应范围内的情况：

#Find values that are in range
in_range = [lo_lim1 <= v1 <= hi_lim1 and lo_lim2 <= v2 <= hi_lim2 for v1, v2 in zip(values1, values2)]

# code is unchanged from here
#Find runs of in-range values
runs = [sum(1 for _ in group) for v, group in groupby(in_range) if v]  # this is the same as yours

#Estimate total time spent in-range
total_time = sum(v if v > 1 else (Buffer_Value*sample_rate) for v in runs)

在你的情况下，如果你使用python 2.x，你可以使用itertools.izip而不是zip来节省一些内存，而对于python 2.x和3.x你可以使用生成器表达式以节省更多：

#Find values that are in range
in_range = (lo_lim1 <= v1 <= hi_lim1 and lo_lim2 <= v2 <= hi_lim2 for v1, v2 in zip(values1, values2))

#Find runs of in-range values
runs = (sum(1 for _ in group) for v, group in groupby(in_range) if v)  # this is the same as yours

#Estimate total time spent in-range
defval = Buffer_Value*sample_rate
total_time = sum(v if v > 1 else defval for v in runs)

两个数据集的python生成器表达式

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