检查 pandas 数据框中的值是否在另一个数据框中其他两个列的任意两个值内

Question

我有两个不同长度的数据帧。 dfSamples（63012375 行）和 dfFixations（200000 行）。

dfSamples = pd.DataFrame({'tSample':[4, 6, 8, 10, 12, 14]})  
dfFixations = pd.DataFrame({'tStart':[4,12],'tEnd':[8,14]})

我想检查 dfSamples 中的每个值是否在 dfFixations 中给出的任意两个范围内，然后为该值分配一个标签。我发现了这个：，但是循环解决方案非常慢，我无法使用任何其他解决方案。

工作（但很慢）示例：

labels = np.empty_like(dfSamples['tSample']).astype(np.chararray)
for i, fixation in dfFix.iterrows():
    log_range = dfSamples['tSample'].between(fixation['tStart'], fixation['tEnd'])
    labels[log_range] = 'fixation'
labels[labels != 'fixation'] = 'no_fixation'
dfSamples['labels'] = labels

按照这个例子：我试图对此进行矢量化，但没有成功。

def check_range(samples, tstart, tend):
    log_range = (samples > tstart) & (samples < tend)
    return log_range
fixations = list(map(check_range, dfSamples['tSample'], dfFix['tStart'], dfFix['tEnd']))

非常感谢任何帮助！

Answer 1

设置

dfSamples = pd.DataFrame({'tSample':[4, 6, 8, 10, 12, 14]})  
dfFixations = pd.DataFrame({'tStart':[4,12],'tEnd':[8,14]})

解决方案

从起点和终点创建区间索引

ii = pd.IntervalIndex.from_arrays(dfFixations['tStart'], dfFixations['tEnd'], closed="both")

ii.contains是一种检查区间索引中的每个区间是否包含一个点的方法，例如

dfSamples["tSample"].apply(ii.contains)

给予

0     [True, False]
1     [True, False]
2     [True, False]
3    [False, False]
4     [False, True]
5     [False, True]
Name: tSample, dtype: object

我们将采用此结果，将 any 函数应用于每个元素（列表）以获得 pandas.Series 个布尔值，然后我们可以将其与 numpy.where 一起使用

dfSamples["labels"] =  np.where(dfSamples["tSample"].apply(ii.contains).apply(any), "fixation", "no_fixation")

结果

   tSample       labels
0        4     fixation
1        6     fixation
2        8  no_fixation
3       10  no_fixation
4       12     fixation
5       14  no_fixation

编辑：更快的版本

使用 piso v0.6.0

import piso
import numpy as np

ii = pd.IntervalIndex.from_arrays(dfFixations['tStart'], dfFixations['tEnd'], closed="both")
contained = np.logical_or.reduce(piso.contains(ii, dfSamples["tSample"], include_index=False), axis=0)
dfSamples["labels"] = np.where(contained, "fixation", "no_fixation")

这将在与@jezrael 的解决方案相似的时间内运行，但是它可以处理间隔重叠的情况，例如

dfFixations = pd.DataFrame({'tStart':[4,5,12],'tEnd':[8,9,14]})

Answer 2

使用IntervalIndex.from_arrays with IntervalIndex.get_indexer , if not match is returned -1, so checked and set ouput in numpy.where:

i = pd.IntervalIndex.from_arrays(dfFixations['tStart'],
                                 dfFixations['tEnd'], 
                                 closed="both")
pos = i.get_indexer(dfSamples['tSample'])
dfSamples['labels'] = np.where(pos != -1, "fixation", "no_fixation")

print (dfSamples)
   tSample       labels
0        4     fixation
1        6     fixation
2        8     fixation
3       10  no_fixation
4       12     fixation
5       14     fixation

性能: 理想中nice sorted not overlap data，实际应该是性能不同，最好测试一下。

dfSamples = pd.DataFrame({'tSample':range(10000)})  
dfFixations = pd.DataFrame({'tStart':range(0, 10000, 5),'tEnd':range(2, 10000, 5)})
    


In [165]: %%timeit
     ...: labels = np.empty_like(dfSamples['tSample']).astype(np.chararray)
     ...: for i, fixation in dfFixations.iterrows():
     ...:     log_range = dfSamples['tSample'].between(fixation['tStart'], fixation['tEnd'])
     ...:     labels[log_range] = 'fixation'
     ...: labels[labels != 'fixation'] = 'no_fixation'
     ...: dfSamples['labels'] = labels
     ...: 
     ...: 
1.25 s ± 52.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)


In [168]: %%timeit
     ...: ii = pd.IntervalIndex.from_arrays(dfFixations['tStart'], dfFixations['tEnd'], closed="both")
     ...: dfSamples["labels1"] =  np.where(dfSamples["tSample"].apply(ii.contains).apply(any), "fixation", "no_fixation")
     ...: 
315 ms ± 18.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)


In [170]: %%timeit
     ...: ii = pd.IntervalIndex.from_arrays(dfFixations['tStart'], dfFixations['tEnd'], closed="both")
     ...: contained = np.logical_or.reduce(piso.contains(ii, dfSamples["tSample"], include_index=False), axis=0)
     ...: dfSamples["labels1"] = np.where(contained, "fixation", "no_fixation")
     ...: 
82.4 ms ± 213 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [166]: %%timeit
     ...: s = pd.IntervalIndex.from_arrays(dfFixations['tStart'], dfFixations['tEnd'], closed="both")
     ...: pos = s.get_indexer(dfSamples['tSample'])
     ...: dfSamples['labels'] = np.where(pos != -1, "fixation", "no_fixation")
     ...: 
27.8 ms ± 1.51 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

检查 pandas 数据框中的值是否在另一个数据框中其他两个列的任意两个值内

Check if value in pandas dataframe is within any two values of two other columns in another dataframe

python

vectorization

dataframe

pandas