英文:
Numpy: How to check if a number is the minimum/maximum among the previous K numbers?
问题
我尝试自动化一个交易策略，该策略应在当前价格是前k个价格中的最小值/最大值时进入/退出多头头寸。
结果应包含以下内容：如果当前数字是前k个数字中的最大值，则为1，如果是最小值则为-1，如果没有条件成立则为0。
例如，如果k = 3，numpy数组 = [1, 2, 3, 2, 1, 6]，结果应该是一个类似的数组：
[0, 0, 1, 0, -1, 1]。
英文:
I'm trying to automate a trading strategy which should enter/exit a long position when the current price is the minimum/maximum among the previous k prices.
The result should contain 1 if the current number is maximum among previous k numbers, -1 if it is the minimum and 0 if none of the conditions are true.
For example if k = 3 and the numpyp array = [1, 2, 3, 2, 1, 6], the result should be an array like:
[0, 0, 1, 0, -1, 1].
I tried the numpy's max function but don't know how to take into account the previous k numbers instead of fixed index and how to switch to default condition for the first k - 1 numbers which should be 0 since there are not k number available to compare them with.
答案1
得分: 4
import pandas as pd
array = [1, 2, 3, 2, 1, 6]
df = pd.DataFrame(array)
df['rolling_max'] = df[0].rolling(3).max()
df['rolling_min'] = df[0].rolling(3).min()
df['result'] = df.apply(lambda row: 1 if row[0] == row['rolling_max'] else (-1 if row[0] == row['rolling_min'] else 0), axis=1)
英文:
I will use Pandas
import pandas as pdarray = [1, 2, 3, 2, 1, 6]df = pd.DataFrame(array)df['rolling_max'] = df[0].rolling(3).max()df['rolling_min'] = df[0].rolling(3).min()df['result'] = df.apply(lambda row: 1 if row[0] == row['rolling_max'] else (-1 if row[0] == row['rolling_min'] else 0), axis=1)
答案2
得分: 1
以下是使用numpy的解决方案，使用了numpy.lib.stride_tricks.sliding_window_view介绍的方法，该方法在版本1.20.0中引入。
注意，这个解决方案（与@Hanwei Tang提出的解决方案一样）并不完全符合您的要求，因为在第二个窗口（[2, 3, 2]）中，2是最小值，因此返回-1而不是您要求的零。但也许您应该重新考虑是否真的希望在第二个窗口中得到零，还是-1。
编辑：如果一个窗口只包含相同的数字，即最小值和最大值相同，这个方法返回零。
import numpy as npdef rolling_max(a, wsize):windows = np.lib.stride_tricks.sliding_window_view(a, wsize)return np.max(windows, axis=-1)def rolling_min(a, wsize):windows = np.lib.stride_tricks.sliding_window_view(a, wsize)return np.min(windows, axis=-1)def check_prize(a, wsize):rmax = rolling_max(a, wsize)rmin = rolling_min(a, wsize)ismax = np.where(a[wsize-1:] == rmax, 1, 0)ismin = np.where(a[wsize-1:] == rmin, -1, 0)result = np.zeros_like(a)result[wsize-1:] = ismax + isminreturn resulta = np.array([1, 2, 3, 2, 1, 6])check_prize(a, wsize=3)# 输出:# array([ 0, 0, 1, -1, -1, 1])b = np.array([1, 2, 4, 3, 1, 6])check_prize(b, wsize=3)# 输出:# array([ 0, 0, 1, 0, -1, 1])c = np.array([1, 2, 2, 2, 1, 6])check_prize(c, wsize=3)# 输出:# array([ 0, 0, 1, 0, -1, 1])
英文:
Here is a solution with numpy using numpy.lib.stride_tricks.sliding_window_view, which was introduced in version 1.20.0.
Note that this solution (like the one proposed by @Hanwei Tang) does not exactly yield the result you was looking for, because in the second window ([2, 3, 2]) 2 is the minimum value and thus a -1 is returned instead of zero (what you requested). But maybe you should rethink whether you really want a zero for the second window or a -1.
EDIT: If a windows only contains same numbers, i.e. the minimum and maximum are the same, this method returns a zero.
import numpy as npdef rolling_max(a, wsize):windows = np.lib.stride_tricks.sliding_window_view(a, wsize)return np.max(windows, axis=-1)def rolling_min(a, wsize):windows = np.lib.stride_tricks.sliding_window_view(a, wsize)return np.min(windows, axis=-1)def check_prize(a, wsize):rmax = rolling_max(a, wsize)rmin = rolling_min(a, wsize)ismax = np.where(a[wsize-1:] == rmax, 1, 0)ismin = np.where(a[wsize-1:] == rmin, -1, 0)result = np.zeros_like(a)result[wsize-1:] = ismax + isminreturn resulta = np.array([1, 2, 3, 2, 1, 6])check_prize(a, wsize=3)# Output:# array([ 0, 0, 1, -1, -1, 1])b = np.array([1, 2, 4, 3, 1, 6])check_prize(b, wsize=3)# Output:# array([ 0, 0, 1, 0, -1, 1])c = np.array([1, 2, 2, 2, 1, 6])check_prize(c, wsize=3)# Output:# array([ 0, 0, 1, 0, -1, 1])
答案3
得分: 1
以下是翻译好的部分：
"Another approach using sliding_window_view with pad:"
from numpy.lib.stride_tricks import sliding_window_view as swvk = 3a = np.array([1, 2, 3, 2, 1, 6])# create sliding windowv = swv(np.pad(a.astype(float), (k-1, 0), constant_values=np.nan), k)# compare each element to min/max of sliding windowout = np.select([np.max(v, 1)==a, np.min(v, 1)==a], [1, -1], 0)
Output: array([ 0, 0, 1, -1, -1, 1])
英文:
Another approach using sliding_window_view with pad:
from numpy.lib.stride_tricks import sliding_window_view as swvk = 3a = np.array([1, 2, 3, 2, 1, 6])# create sliding windowv = swv(np.pad(a.astype(float), (k-1, 0), constant_values=np.nan), k)# compare each element to min/max of sliding windowout = np.select([np.max(v, 1)==a, np.min(v, 1)==a], [1, -1], 0)
Output: array([ 0, 0, 1, -1, -1, 1])
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