我想以仅连续的方式过滤 numpy
array
(或 pandas
DataFrame
)保留长度至少为 window_size
的一系列相同值,其他所有值都设置为 0。
例如:
[1,1,1,0,0,1,1,1,1,0,0,1,0,0,0,1,1,1,0,1,1,1,1]
当使用 4 的窗口大小时应该变成
[0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,0,0,1,1,1,1]
我尝试过使用 rolling_apply
和 scipy.ndimage.filtes.gerneric_filter
但由于滚动内核函数的性质,我认为这不是正确的方法在这里(我现在坚持使用它)。
无论如何,我在这里插入我的尝试:
import numpy as np
import pandas as pd
import scipy
#from scipy import ndimage
df= pd.DataFrame({'x':np.array([1,1,1,0,0,1,1,1,1,0,0,1,0,0,0,1,1,1,0,1,1,1,1])})
df_alt = df.copy()
def filter_df(df, colname, window_size):
rolling_func = lambda z: z.sum() >= window_size
df[colname] = pd.rolling_apply(df[colname],
window_size,
rolling_func,
min_periods=window_size/2,
center = True)
def filter_alt(df, colname, window_size):
rolling_func = lambda z: z.sum() >= window_size
return scipy.ndimage.filters.generic_filter(df[colname].values,
rolling_func,
size = window_size,
origin = 0)
window_size = 4
filter_df(df, 'x', window_size)
print df
filter_alt(df_alt, 'x', window_size)
最佳答案
这基本上是一个 image closing operation in image-processing
但是对于一维案例。这样的操作可以用卷积方法来实现。现在,NumPy does support 1D convolution
,所以我们很幸运!因此,要解决我们的问题,它会是这样的 -
def conv_app(A, WSZ):
K = np.ones(WSZ,dtype=int)
L = WSZ-1
return (np.convolve(np.convolve(A,K)>=WSZ,K)[L:-L]>0).astype(int)
sample 运行-
In [581]: A
Out[581]: array([1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1])
In [582]: conv_app(A,4)
Out[582]: array([0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1])
In [583]: A = np.append(1,A) # Append 1 and see what happens!
In [584]: A
Out[584]: array([1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1])
In [585]: conv_app(A,4)
Out[585]: array([1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1])
运行时测试 -
本节对列出的解决已发布问题的其他几种方法进行了基准测试。它们的定义如下 -
def groupby_app(A,WSZ): # @lambo477's solution
groups = itertools.groupby(A)
result = []
for group in groups:
group_items = [item for item in group[1]]
group_length = len(group_items)
if group_length >= WSZ:
result.extend([item for item in group_items])
else:
result.extend([0]*group_length)
return result
def stride_tricks_app(arr, window): # @ajcr's solution
x = pd.rolling_min(arr, window)
x[:window-1] = 0
y = np.lib.stride_tricks.as_strided(x, (len(x)-window+1, window), (8, 8))
y[y[:, -1] == 1] = 1
return x.astype(int)
时间 -
In [541]: A = np.random.randint(0,2,(100000))
In [542]: WSZ = 4
In [543]: %timeit groupby_app(A,WSZ)
10 loops, best of 3: 74.5 ms per loop
In [544]: %timeit stride_tricks_app(A,WSZ)
100 loops, best of 3: 3.35 ms per loop
In [545]: %timeit conv_app(A,WSZ)
100 loops, best of 3: 2.82 ms per loop
关于python - 为具有最小窗口长度的连续系列过滤 pandas 或 numpy 数组,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34616439/