Python 从keras数据的数据帧生成numpy数据数组
这是我一直在思考的一项任务。我有一个包含用户运动特征的数据帧(按用户Python 从keras数据的数据帧生成numpy数据数组,python,pandas,keras,deep-learning,numpy-ndarray,Python,Pandas,Keras,Deep Learning,Numpy Ndarray,这是我一直在思考的一项任务。我有一个包含用户运动特征的数据帧(按用户id),类似于下面的数据帧: >>> df id speed1 speed2 acc1 acc2 label 0 1 19 12 5 2 0 1 1 10 11 9 3 0 2 1 12 10 4 -1 0 3 1 29 13
id
),类似于下面的数据帧:
>>> df
id speed1 speed2 acc1 acc2 label
0 1 19 12 5 2 0
1 1 10 11 9 3 0
2 1 12 10 4 -1 0
3 1 29 13 8 4 0
4 1 30 23 9 10 0
5 1 18 11 2 -1 0
6 1 10 6 -3 -2 0
7 2 5 1 0 0 1
8 2 7 2 1 3 1
9 2 6 2 1 0 1
从这个数据帧中,我想通过分割每个用户的(即id
)记录来生成一个固定长度段的numpy ndarray
(我应该说数组列表吗?),这样每个段的形状都是(1,5,4)
,我可以通过以下方式将其输入神经网络:
- 每个段(因此,
)由上述数据帧中运动特性1
(因此,speed1 speed2 acc1 acc2
)的五个数组(因此,4
)组成5
- 如果行不能组成五个数组,则剩余的数组用零填充(即零填充)
标签
列也应该是一个单独的数组,通过在填充段的零填充数组位置复制标签
的值来匹配新数组的大小
在上述给定的df
示例中,预期输出为:
>>>input_array
[
[
[19 12 5 2]
[10 11 9 3]
[12 10 4 -1]
[29 13 8 4]
[30 23 9 10]
]
[
[18 11 2 -1]
[10 6 -3 -2]
[0 0 0 0]
[0 0 0 0]
[0 0 0 0]
]
[
[5 6 -3 -2]
[7 2 1 3]
[6 2 1 0]
[0 0 0 0]
[0 0 0 0]
]
]
有7行,因此最后3行是零填充的。类似地,id=1
有3行,因此最后2行是零填充的id=2
df2 = {
'id': [1,1,1,1,1,1,1,1,1,1,1,1],
'speed1': [17.63,17.63,0.17,1.41,0.61,0.32,0.18,0.43,0.30,0.46,0.75,0.37],
'speed2': [0.00,-0.09,1.24,-0.80,-0.29,-0.14,0.25,-0.13,0.16,0.29,-0.38,0.27],
'acc1': [0.00,0.01,-2.04,0.51,0.15,0.39,-0.38,0.29,0.13,-0.67,0.65,0.52],
'acc2': [29.03,56.12,18.49,11.85,36.75,27.52,81.08,51.06,19.85,10.76,14.51,24.27],
'label' : [3,3,3,3,3,3,3,3,3,3,3,3] }
df2 = pd.DataFrame.from_dict(df2)
X , y = transform(df2[:10])
X
array([[[[ 1.763e+01, 0.000e+00, 0.000e+00, 2.903e+01],
[ 1.763e+01, -9.000e-02, 1.000e-02, 5.612e+01],
[ 1.700e-01, 1.240e+00, -2.040e+00, 1.849e+01],
[ 1.410e+00, -8.000e-01, 5.100e-01, 1.185e+01],
[ 6.100e-01, -2.900e-01, 1.500e-01, 3.675e+01]]],
[[[ 0.000e+00, 0.000e+00, 0.000e+00, 0.000e+00],
[ 0.000e+00, 0.000e+00, 0.000e+00, 0.000e+00],
[ 0.000e+00, 0.000e+00, 0.000e+00, 0.000e+00],
[ 0.000e+00, 0.000e+00, 0.000e+00, 0.000e+00],
[ 0.000e+00, 0.000e+00, 0.000e+00, 0.000e+00]]],
[[[ 3.200e-01, -1.400e-01, 3.900e-01, 2.752e+01],
[ 1.800e-01, 2.500e-01, -3.800e-01, 8.108e+01],
[ 4.300e-01, -1.300e-01, 2.900e-01, 5.106e+01],
[ 3.000e-01, 1.600e-01, 1.300e-01, 1.985e+01],
[ 4.600e-01, 2.900e-01, -6.700e-01, 1.076e+01]]]])
请注意函数如何引入一个全零数组作为第二个元素。理想情况下,输出应该只包含第一个和最后一个数组
索引不能包含负值的错误
df2
你会得到:
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-71-743489875901> in <module>()
----> 1 X , y = transform(df2)
2 X
2 frames
<ipython-input-55-f6e028a2e8b8> in transform(dataframe, chunk_size)
24 inpt = np.pad(
25 inpt, [(0, chunk_size-len(inpt)),(0, 0)],
---> 26 mode='constant')
27 # add each inputs split to accumulators
28 X = np.concatenate([X, inpt[np.newaxis, np.newaxis]], axis=0)
<__array_function__ internals> in pad(*args, **kwargs)
/usr/local/lib/python3.6/dist-packages/numpy/lib/arraypad.py in pad(array, pad_width, mode, **kwargs)
746
747 # Broadcast to shape (array.ndim, 2)
--> 748 pad_width = _as_pairs(pad_width, array.ndim, as_index=True)
749
750 if callable(mode):
/usr/local/lib/python3.6/dist-packages/numpy/lib/arraypad.py in _as_pairs(x, ndim, as_index)
517
518 if as_index and x.min() < 0:
--> 519 raise ValueError("index can't contain negative values")
520
521 # Converting the array with `tolist` seems to improve performance
ValueError: index can't contain negative values
---------------------------------------------------------------------------
ValueError回溯(最近一次调用上次)
在()
---->1 X,y=变换(df2)
2 X
2帧
转换中(数据帧、块大小)
24输入=np.pad(
25输入,[(0,块大小透镜(输入)),(0,0)],
--->26模式(常数)
27#将每个分离的输入添加到累加器
28 X=np.连接([X,输入[np.newaxis,np.newaxis]],轴=0)
在pad(*args,**kwargs)中
/pad中的usr/local/lib/python3.6/dist-packages/numpy/lib/arraypad.py(数组、pad_宽度、模式,**kwargs)
746
747#广播成型(array.ndim,2)
-->748 pad_width=_as_pairs(pad_width,array.ndim,as_index=True)
749
750如果可调用(模式):
/usr/local/lib/python3.6/dist-packages/numpy/lib/arraypad.py成对(x,ndim,as_索引)
517
518如果as_索引和x.min()<0:
-->519提升值错误(“索引不能包含负值”)
520
521#使用“tolist”转换数组似乎可以提高性能
ValueError:索引不能包含负值
[编辑]修复了错误。下面的实现现在应该提供所需的输出:
import pandas as pd
import numpy as np
df = {
'id': [1,1,1,1,1,1,1,1,1,1,1,1],
'speed1': [17.63,17.63,0.17,1.41,0.61,0.32,0.18,0.43,0.30,0.46,0.75,0.37],
'speed2': [0.00,-0.09,1.24,-0.80,-0.29,-0.14,0.25,-0.13,0.16,0.29,-0.38,0.27],
'acc1': [0.00,0.01,-2.04,0.51,0.15,0.39,-0.38,0.29,0.13,-0.67,0.65,0.52],
'acc2': [29.03,56.12,18.49,11.85,36.75,27.52,81.08,51.06,19.85,10.76,14.51,24.27],
'label' : [3,3,3,3,3,3,3,3,3,3,3,3] }
df = pd.DataFrame.from_dict(df)
def transform(dataframe, chunk_size=5):
grouped = dataframe.groupby('id')
# initialize accumulators
X, y = np.zeros([0, 1, chunk_size, 4]), np.zeros([0,])
# loop over each group (df[df.id==1] and df[df.id==2])
for _, group in grouped:
inputs = group.loc[:, 'speed1':'acc2'].values
label = group.loc[:, 'label'].values[0]
# calculate number of splits
N = (len(inputs)-1) // chunk_size
if N > 0:
inputs = np.array_split(
inputs, [chunk_size + (chunk_size*i) for i in range(N)])
else:
inputs = [inputs]
# loop over splits
for inpt in inputs:
inpt = np.pad(
inpt, [(0, chunk_size-len(inpt)),(0, 0)],
mode='constant')
# add each inputs split to accumulators
X = np.concatenate([X, inpt[np.newaxis, np.newaxis]], axis=0)
y = np.concatenate([y, label[np.newaxis]], axis=0)
return X, y
X, y = transform(df)
print('X shape =', X.shape)
print('X =', X)
print('Y shape =', y.shape)
print('Y =', y)
# >> out:
# X shape = (3, 1, 5, 4)
# X = [[[[17.63 0. 0. 29.03]
# [17.63 -0.09 0.01 56.12]
# [ 0.17 1.24 -2.04 18.49]
# [ 1.41 -0.8 0.51 11.85]
# [ 0.61 -0.29 0.15 36.75]]]
#
#
# [[[ 0.32 -0.14 0.39 27.52]
# [ 0.18 0.25 -0.38 81.08]
# [ 0.43 -0.13 0.29 51.06]
# [ 0.3 0.16 0.13 19.85]
# [ 0.46 0.29 -0.67 10.76]]]
#
#
# [[[ 0.75 -0.38 0.65 14.51]
# [ 0.37 0.27 0.52 24.27]
# [ 0. 0. 0. 0. ]
# [ 0. 0. 0. 0. ]
# [ 0. 0. 0. 0. ]]]]
# Y shape = (3,)
# Y = [3. 3. 3.]
我已经更新了上面的代码(输出现在是(3,1,5,4)),谢谢。准备数据输入到conv网络,目标是将输入转换为网格状形状
(1100,4)
100个数据点(宽度)的1段(长度)和4个运动特征((通道)。我会尽力提供帮助。注意:我已经更新了两次答案/实现。在第一次更新之后,它仍然存在一些问题/bug,但在第二次更新之后,现在应该可以了。如何使用np.array\u split
(数组在错误的位置被拆分),导致chunk\u size-len(inpt)进行拆分存在问题
为负值(传递给np.pad
的参数pad\u width
(第二个参数),提高了ValueError
)。