Question

我有一个大数据集（＆gt; 200k），我试图用一个值替换零序列。具有2个以上零的零序列是一个工件，应通过将其设置为np.NAN来删除。

我已阅读Searching a sequence in a NumPy array但它并不完全符合我的要求，因为我没有静态模式。

np.array([0, 1.0, 0, 0, -6.0, 13.0, 0, 0, 0, 1.0, 16.0, 0, 0, 0, 0, 1.0, 1.0, 1.0, 1.0])
# should be converted to this
np.array([0, 1.0, 0, 0, -6.0, 13.0, NaN, NaN, NaN, 1.0, 16.0, NaN, NaN, NaN, NaN, 1.0, 1.0, 1.0, 1.0])

如果您需要更多信息，请与我们联系。提前谢谢！

<小时/> 结果：

感谢您的回答，这是我的（非专业）测试结果在288240点运行

divakar took 0.016000ms to replace 87912 points
desiato took 0.076000ms to replace 87912 points
polarise took 0.102000ms to replace 87912 points

因为@Divakar的解决方案是最短也是最快的我接受他的解决方案。

Answer 1

那基本上是一个binary closing operation，对收盘差距有一个门槛要求。这是基于它的实现 -

# Pad with ones so as to make binary closing work around the boundaries too
a_extm = np.hstack((True,a!=0,True))

# Perform binary closing and look for the ones that have not changed indiicating
# the gaps in those cases were above the threshold requirement for closing
mask = a_extm == binary_closing(a_extm,structure=np.ones(3))

# Out of those avoid the 1s from the original array and set rest as NaNs
out = np.where(~a_extm[1:-1] & mask[1:-1],np.nan,a)

一种方法可以避免在前面的方法中根据需要使用边界元素，这可能会使处理大型数据集时有点昂贵，就像这样 -

# Create binary closed mask
mask = ~binary_closing(a!=0,structure=np.ones(3))
idx = np.where(a)[0]
mask[:idx[0]] = idx[0]>=3
mask[idx[-1]+1:] = a.size - idx[-1] -1 >=3

# Use the mask to set NaNs in a
out = np.where(mask,np.nan,a)

Answer 2

以下是您可以用于列表的功能：

import numpy as np

def replace(a_list):
    for i in xrange(len(a_list) - 2):
        print a_list[i:i+3]
        if (a_list[i] == 0 and a_list[i+1] == 0 and a_list[i+2] == 0) or (a_list[i] is np.NaN and a_list[i+1] is np.NaN and a_list[i+2] == 0):
            a_list[i] = np.NaN
            a_list[i+1] = np.NaN
            a_list[i+2] = np.NaN
    return a_list

由于列表是在一个方向上遍历的，因此您只需进行两次比较：(0, 0, 0)或(NaN, NaN, 0)，因为您将0替换为NaN。

Answer 3

您可以使用groupby包的itertools

resolve

给你

import numpy as np
from itertools import groupby

l = np.array([0, 1, 0, 0, -6, 13, 0, 0, 0, 1, 16, 0, 0, 0, 0])

def _ret_list( k, it ):
    # number of elements in iterator, i.e., length of list of similar items
    l = sum( 1 for i in it )

    if k==0 and l>2:
        # sublist has more than two zeros. replace each zero by np.nan
        return [ np.nan ]*l
    else:
        # return sublist of simliar items
        return [ k ]*l

# group items and apply _ret_list on each group
procesed_l = [_ret_list(k,g) for k,g in groupby(l)]
# flatten the list and convert to a numpy array
procesed_l = np.array( [ item for l in procesed_l for item in l ] )

print procesed_l

请注意，每个[ 0. 1. 0. 0. -6. 13. nan nan nan 1. 16. nan nan nan nan]都会转换为int。见这里：NumPy or Pandas: Keeping array type as integer while having a NaN value

用其他值替换零序列

3 个答案: