Python 如何按间隔索引分组,聚合列表上的平均值,并连接到另一个数据帧?
我有两个数据帧。它们看起来像这样:Python 如何按间隔索引分组,聚合列表上的平均值,并连接到另一个数据帧?,python,pandas,numpy,Python,Pandas,Numpy,我有两个数据帧。它们看起来像这样: df_a Framecount probability 0 0.0 [0.00019486549333333332, 4.883635666666667e-06... 1 1.0 [0.00104359155, 3.9232405e-05, 0.0015722045000... 2 2.0 [0.000485
df_a
Framecount probability
0 0.0 [0.00019486549333333332, 4.883635666666667e-06...
1 1.0 [0.00104359155, 3.9232405e-05, 0.0015722045000...
2 2.0 [0.00048501002666666667, 1.668179e-05, 0.00052...
3 3.0 [4.994969500000001e-05, 4.0931635e-07, 0.00011...
4 4.0 [0.0004808829, 5.389742e-05, 0.002522127933333...
.. ... ...
906 906.0 [1.677140566666667e-05, 1.1745095666666665e-06...
907 907.0 [1.5164155000000002e-05, 7.66629575e-07, 0.000...
908 908.0 [8.1334184e-05, 0.00012675669636333335, 0.0028...
909 909.0 [0.00014893802999999998, 1.0407592500000001e-0...
910 910.0 [4.178489e-05, 2.17477925e-06, 0.02094931, 0.0...
以及:
当df_a.Framecount
介于df_b.start和df_b.stop之间时,我想将df_a.probability
合并到df_b
。df_a.probability
的聚合统计应该是mean
,但我遇到了错误,因为df_a.probability
是dtype np array
我正在尝试使用以下代码:
idx = pd.IntervalIndex.from_arrays(df_text['start'], df_text['stop'])
df_text.join(df_vid.groupby(idx.get_indexer_non_unique(df_vid['Framecount']))['probability'].apply(np.mean), how='left')
第1行创建索引并确定分组。在第2行中,我试图实现group by,并将df_a.probability
中属于groupby平均指数的所有值进行聚合。我希望每个groupby有一个数组,它是groupby索引中所有数组的平均值。此代码给出了以下错误:
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-271-19c7d58fb664> in <module>
1 idx = pd.IntervalIndex.from_arrays(df_text['start'], df_text['stop'])
2 f = lambda x: np.mean(np.array(x.tolist()), axis=0)
----> 3 df_text.join(df_vid.groupby(idx.get_indexer_non_unique(df_vid['Framecount']))['probability'].apply(np.mean), how='left')
~/anaconda3/lib/python3.7/site-packages/pandas/core/frame.py in groupby(self, by, axis, level, as_index, sort, group_keys, squeeze, observed)
5808 group_keys=group_keys,
5809 squeeze=squeeze,
-> 5810 observed=observed,
5811 )
5812
~/anaconda3/lib/python3.7/site-packages/pandas/core/groupby/groupby.py in __init__(self, obj, keys, axis, level, grouper, exclusions, selection, as_index, sort, group_keys, squeeze, observed, mutated)
407 sort=sort,
408 observed=observed,
--> 409 mutated=self.mutated,
410 )
411
~/anaconda3/lib/python3.7/site-packages/pandas/core/groupby/grouper.py in get_grouper(obj, key, axis, level, sort, observed, mutated, validate)
588
589 elif is_in_axis(gpr): # df.groupby('name')
--> 590 if gpr in obj:
591 if validate:
592 obj._check_label_or_level_ambiguity(gpr, axis=axis)
~/anaconda3/lib/python3.7/site-packages/pandas/core/generic.py in __contains__(self, key)
1848 def __contains__(self, key) -> bool_t:
1849 """True if the key is in the info axis"""
-> 1850 return key in self._info_axis
1851
1852 @property
~/anaconda3/lib/python3.7/site-packages/pandas/core/indexes/base.py in __contains__(self, key)
3898 @Appender(_index_shared_docs["contains"] % _index_doc_kwargs)
3899 def __contains__(self, key) -> bool:
-> 3900 hash(key)
3901 try:
3902 return key in self._engine
TypeError: unhashable type: 'numpy.ndarray'
或
或
我得到了同样的错误
如何做到这一点?我希望有人能想出一个不涉及循环的解决方案,但由于所有内容都已排序,我认为性能实际上不会那么差(两个数据帧的长度呈线性,没有内存开销)。我不知道您的数据帧的精确规格,因此我将首先创建一些示例
n_a = 11
df_a = pd.DataFrame(
{"Framecount": list(range(n_a)), "probability": np.random.rand(n_a)}
)
n_b = 6
start = np.linspace(0, n_a, n_b)
end = start + n_a / (n_b - 1) - 1e-5
df_b = pd.DataFrame({"start": start, "end": end, "mean": [np.nan] * n_b})
print(df_a)
Framecount probability
0 0 0.099412
1 1 0.492661
2 2 0.043000
3 3 0.382923
4 4 0.208177
5 5 0.110007
6 6 0.369756
7 7 0.324723
8 8 0.702838
9 9 0.182167
10 10 0.578837
print(df_b)
start end mean
0 0.0 2.19999 NaN
1 2.2 4.39999 NaN
2 4.4 6.59999 NaN
3 6.6 8.79999 NaN
4 8.8 10.99999 NaN
5 11.0 13.19999 NaN
现在我将循环遍历数据帧,聚合当前开始
和结束
之间的所有值,并在dfu b
中的相应行中赋值:
i = j = 0
while i < n_a and j < n_b:
# seek to next row of df_b where start <= df_a[i]
while i < n_a and df_a.loc[i, "Framecount"] < df_b.loc[j, "start"]:
i += 1
accum = 0
count = 0
while i < n_a and df_a.loc[i, "Framecount"] < df_b.loc[j, "end"]:
accum += df_a.loc[i, "probability"]
count += 1
i += 1
df_b.loc[j, "mean"] = accum / count
j += 1
print(df_b)
start end mean
0 0.0 2.19999 0.211691
1 2.2 4.39999 0.29555
2 4.4 6.59999 0.239882
3 6.6 8.79999 0.513781
4 8.8 10.99999 0.380502
5 11.0 13.19999 NaN
i=j=0
而i
出现此错误的原因是idx.get\u indexer\u non\u unique(df\u vid['Framecount'])
创建了一个元组
,而您不能以这种方式groupby
元组。
df_vid.groupby(idx.get_indexer\u non_unique(df_vid['Framecount'])[0])
选择元组中的第一个数组将有效
idx.get\u indexer(df\u a.fc)
将生成一个数组,其索引为fc
所属的间隔。如果没有匹配的时间间隔,索引将显示为-1
df_a.groupby(idx.get_indexer(df_a.fc))
按索引数组分组
.agg({'prob':list})
将每个fc
的所有列表聚合到一个列表中。
- 每个组的结果是一个列表列表
.prob.map(np.mean)
返回组中所有列表的总体平均值
.prob.apply(λx:[np.mean(v)表示x中的v])
为每个列表返回一个平均值列表
没有'fc'
值落入12.12-12.47的存储箱中
将熊猫作为pd导入
将numpy作为np导入
#使用开始和停止范围设置df
数据={'start':[12.12,13.44,20.88,31.61,33.44,880.44,888.63,892.13895.31907.58],'stop':[12.47,20.82,29.63,33.33,42.21887.92,892.07895.3900.99,908.35]}
df=pd.DataFrame(数据)
#设置样本df_a,帧数为fc,概率为prob
np.random.seed(365)
df_a=pd.DataFrame({'fc':range(911),'prob':np.random.randint(1100,(911,14)).tolist())
#这将把列转换为np.array而不是list;代码的其余部分无论如何都可以工作
#df_a.prob=df_a.prob.map(np.array)
#从df start和stop创建IntervalIndex
idx=pd.IntervalIndex.from_数组(df.start、df.stop、closed='both')
这将在轴=0上创建一个平均值列表
dfg=df_a.groupby(idx.get_indexer(df_a.fc)).agg({'prob':list}).prob.apply(lambda x:np.mean(x,axis=0))
#将df与dfg连接起来
dfj=df.join(dfg)
#方法列表的显示(dfj)
启停探头
0 12.12 12.47南
1 13.44 20.82 [49.3, 57.1, 51.4, 45.9, 47.1, 45.9, 45.9, 55.3, 32.6, 48.0, 42.0, 45.0, 50.4, 54.4]
2 20.88 29.63 [42.7, 42.6, 46.0, 45.9, 54.1, 55.9, 50.1, 55.2, 51.7, 54.0, 37.6, 60.9, 49.2, 45.6]
3 31.61 33.33 [87.5, 49.0, 46.5, 54.5, 75.0, 47.0, 24.0, 40.5, 52.5, 21.0, 51.0, 72.5, 34.5, 50.5]
4 33.44 42.21 [48.6, 66.2, 45.8, 64.7, 43.1, 69.0, 54.4, 52.1, 52.6, 59.6, 51.1, 42.1, 43.3, 38.0]
5 880.44 887.92 [51.9, 50.6, 63.7, 47.7, 51.3, 34.9, 51.3, 53.0, 53.4, 65.1, 38.6, 49.4, 48.1, 44.1]
6 888.63 892.07 [45.2, 23.5, 67.2, 68.0, 38.2, 47.2, 50.2, 75.8, 35.2, 46.8, 55.0, 57.5, 44.2, 78.0]
7 892.13 895.30 [61.3, 44.0, 43.3, 36.3, 63.7, 89.7, 51.7, 57.0, 50.0, 68.7, 80.7, 46.3, 66.7, 11.3]
8 895.31 900.99 [68.2, 44.6, 50.8, 35.2, 53.2, 40.4, 34.8, 77.4, 61.0, 35.2, 26.0, 47.8, 30.4, 55.4]
9 907.58 908.35 [17.0, 78.0, 24.0, 33.0, 88.0, 3.0, 43.0, 2.0, 36.0, 48.0, 8.0, 87.0, 36.0, 34.0]
这将为每组创建一个平均值
dfg=df_a.groupby(idx.get_indexer(df_a.fc)).agg({'prob':list}).prob.map(np.mean)
#将df与dfg连接起来
dfj=df.join(dfg)
#总平均值显示(dfj)
启停探头
0 12.12 12.47南
1 13.44 20.82 47.877551
2 20.88 29.63 49.380952
3 31.61 33.33 50.428571
4 33.44 42.21 52.182540
5 880.44 887.92 50.224490
6 888.63 892.07 52.303571
7 892.13 895.30 55.047619
8 895.31 900.99 47.171429
9 907.58 908.35 38.357143
感谢您的回复。我很困惑。什么是n_a和n_b?@connor449只是组成数据帧的长度。对你来说,它们都是906。我得到了这个错误:`----------------------------------------------------------------------ZeroDivisionError回溯(最后一次调用)在17 I+=118--->19 df_b.loc[j,“mean”]=累计/计数20 j+=1 21零除误差:零除`
df_text.join(df_vid.groupby(idx.get_indexer_non_unique(df_vid['Framecount']))['probability'].apply((np.mean), how='left')
df_text.join(df_vid.groupby(idx.get_indexer_non_unique(df_vid['Framecount']))['probability'].mean()), how='left')
n_a = 11
df_a = pd.DataFrame(
{"Framecount": list(range(n_a)), "probability": np.random.rand(n_a)}
)
n_b = 6
start = np.linspace(0, n_a, n_b)
end = start + n_a / (n_b - 1) - 1e-5
df_b = pd.DataFrame({"start": start, "end": end, "mean": [np.nan] * n_b})
print(df_a)
Framecount probability
0 0 0.099412
1 1 0.492661
2 2 0.043000
3 3 0.382923
4 4 0.208177
5 5 0.110007
6 6 0.369756
7 7 0.324723
8 8 0.702838
9 9 0.182167
10 10 0.578837
print(df_b)
start end mean
0 0.0 2.19999 NaN
1 2.2 4.39999 NaN
2 4.4 6.59999 NaN
3 6.6 8.79999 NaN
4 8.8 10.99999 NaN
5 11.0 13.19999 NaN
i = j = 0
while i < n_a and j < n_b:
# seek to next row of df_b where start <= df_a[i]
while i < n_a and df_a.loc[i, "Framecount"] < df_b.loc[j, "start"]:
i += 1
accum = 0
count = 0
while i < n_a and df_a.loc[i, "Framecount"] < df_b.loc[j, "end"]:
accum += df_a.loc[i, "probability"]
count += 1
i += 1
df_b.loc[j, "mean"] = accum / count
j += 1
print(df_b)
start end mean
0 0.0 2.19999 0.211691
1 2.2 4.39999 0.29555
2 4.4 6.59999 0.239882
3 6.6 8.79999 0.513781
4 8.8 10.99999 0.380502
5 11.0 13.19999 NaN