我从元组列表开始（每个元组都是一个（X，Y））。我的最终结果是我想使用 numpy 找到长度为 4 的每个窗口/bin 内的最大 Y 值。# List of tuples[(0.05807200929152149, 9.9720125), (0.34843205574912894, 1.1142874), (0.6387921022067363, 2.0234027), (0.9291521486643438, 1.4435122), (1.207897793263647, 2.3677678), (1.4982578397212543, 1.9457655), (1.7886178861788617, 2.8343441), (2.078977932636469, 5.7816567), ...]# Convert to numpy arraydt = np.dtype('float,float')arr = np.asarray(listTuples, dt)# [(0.05807201, 9.97201252) (0.34843206, 1.11428738)# (0.6387921 , 2.02340269) (0.92915215, 1.4435122 )# (1.20789779, 2.36776781) (1.49825784, 1.9457655 )# (1.78861789, 2.83434415) (2.07897793, 5.78165674)# (2.36933798, 3.14842606) ...]#Create windows/blocks of 4 elementsarr = arr.reshape(-1,4)# [[(0.05807201, 9.97201252) (0.34843206, 1.11428738)# (0.6387921 , 2.02340269) (0.92915215, 1.4435122 )]# [(1.20789779, 2.36776781) (1.49825784, 1.9457655 )# (1.78861789, 2.83434415) (2.07897793, 5.78165674)]# [(2.36933798, 3.14842606) (2.95005807, 2.10357308)# (3.24041812, 1.15985966) (3.51916376, 2.03056955)]...]print(arr.max(axis=1)) <-- ERROR HEREprint(max(arr,key=lambda x:x[1])) <-- ERROR, tried this too but doesn't work我想要使用最大 y 值从每个窗口/块获得的预期输出如下。或者，格式可以是常规元组列表，并不严格需要是 numpy 数组：[[(0.05807201, 9.97201252)][(2.07897793, 5.78165674)][(2.36933798, 3.14842606)]...]OR other format:[(0.05807201, 9.97201252),(2.07897793, 5.78165674),(2.36933798, 3.14842606)]...]
3 回答
牧羊人nacy
TA贡献1862条经验 获得超7个赞
这是一种矢量化（可能是最快的）方法：
arr = np.asarray(listTuples, np.dtype('float,float'))
idx = arr.view(('float',2))[:,1].reshape(-1,4).argmax(1)
arr = arr.reshape(-1,4)[np.arange(len(idx)),idx]
#[(0.05807201, 9.9720125) (2.07897793, 5.7816567) ...]
您基本上使用结构化数组的数组（非结构化）版本，并view沿 查找 Y 的 argmax axis=1。然后使用这些索引来过滤原始数组中的元组arr。
HUX布斯
TA贡献1876条经验 获得超6个赞
这应该可以解决你的问题。
输入：元组列表
输出：元组列表，取每个 4 个元素块中 y 值最大的元组 import numpy as np
# List of tuples
listTuples = [(1,1),(120,1000),(12,90),(1,1),(0.05807200929152149, 9.9720125),
(0.34843205574912894, 1.1142874), (0.6387921022067363, 2.0234027),
(0.9291521486643438, 1.4435122), (1.207897793263647, 2.3677678),
(1.4982578397212543, 1.9457655), (1.7886178861788617, 2.8343441),
(2.078977932636469, 5.7816567)]
def extractMaxY(li):
result = []
index = 0
for i in range(0,len(li), 4):
max = -100000
#find the max Y in blocks of 4
for j in range(4):
if li[i+j][1] > max:
max = li[i+j][1]
index = i+j
result.append(li[index])
return result
print(extractMaxY(listTuples))
然后输出是
[(120, 1000), (0.05807200929152149, 9.9720125), (2.078977932636469,
5.7816567)]
应该如此，对吗？
有只小跳蛙
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你可以试试这个
import numpy as np
ans = []
for value, comp in zip(x, np.max(x, axis=1)[:,1]):
ans.append([(i, j) for i, j in value if np.isclose(j,comp)])
print(np.array(ans))
将其应用于您的数据
x = np.array([[(0.05807201, 9.97201252), (0.34843206, 1.11428738),
(0.6387921 , 2.02340269), (0.92915215, 1.4435122 )],
[(1.20789779, 2.36776781), (1.49825784, 1.9457655 ),
(1.78861789, 2.83434415), (2.07897793, 5.78165674)],
[(2.36933798, 3.14842606), (2.95005807, 2.10357308),
(3.24041812, 1.15985966), (3.51916376, 2.03056955)]])
退货
[[[ 0.05807201 9.97201252]]
[[ 2.07897793 5.78165674]]
[[ 2.36933798 3.14842606]]]
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