找到最小值之前的每列值

时间:2016-11-18 06:50:13

标签: python pandas numpy

考虑数组a

a = np.array([
        [5, 4],
        [4, 5],
        [2, 2],
        [6, 1],
        [3, 7]
    ])

我可以找到最小值的位置

a.argmin(0)

array([2, 3])

如何找到索引2之前的第0列的最大值。第1列和索引3的值相同。更重要的是,它们在哪里?

如果我这样做

a.max(0)

array([6, 7])

但我需要

# max values
array([5, 5])

# argmax before mins
array([0, 1])

2 个答案:

答案 0 :(得分:4)

这是使用broadcasting -

的一种方法
b = np.where(a.argmin(0) >= np.arange(a.shape[0])[:,None],a,np.nan)
idx = np.nanargmax(b,axis=0)
out = a[idx,np.arange(a.shape[1])]

示例运行 -

In [38]: a
Out[38]: 
array([[5, 4],
       [4, 5],
       [2, 2],
       [6, 1],
       [3, 7]])

In [39]: b = np.where(a.argmin(0) >= np.arange(a.shape[0])[:,None],a,np.nan)
    ...: idx = np.nanargmax(b,axis=0)
    ...: out = a[idx,np.arange(a.shape[1])]
    ...: 

In [40]: idx
Out[40]: array([0, 1])

In [41]: out
Out[41]: array([5, 5])

或者,如果a仅包含正数,我们只需使用 - {/ p>即可获得idx

mask = a.argmin(0) >= np.arange(a.shape[0])[:,None]
idx = (a*mask).argmax(0)

答案 1 :(得分:2)

我知道我可以使用累积argmax @ajcr answered that question for me here

的矢量化版本来回答这个问题
def ajcr(a):
    m = np.maximum.accumulate(a)
    x = np.repeat(np.arange(a.shape[0])[:, None], a.shape[1], axis=1)
    x[1:] *= m[:-1] < m[1:]
    np.maximum.accumulate(x, axis=0, out=x)
    # at this point x houses the cumulative argmax
    # we slice that with a's argmin
    return x[a.argmin(0), np.arange(a.shape[1])]

def divakar(a):
    b = np.where(a.argmin(0) >= np.arange(a.shape[0])[:,None],a,np.nan)
    return np.nanargmax(b,axis=0)

比较

a = np.random.randn(10000, 1000)
(ajcr(a) == divakar(a)).all()

True

时间

import timeit

results = pd.DataFrame(
    [], [10, 100, 1000, 10000],
    pd.MultiIndex.from_product(
        [['divakar', 'ajcr'], [10, 100, 1000]]))

for i, j in results.stack(dropna=False).index:
    a = np.random.randn(i, j)
    results.loc[i, ('divakar', j)] = \
        timeit.timeit(
            'divakar(a)',
            setup='from __main__ import divakar, a',
            number=10)
    results.loc[i, ('ajcr', j)] = \
        timeit.timeit(
            'ajcr(a)',
            setup='from __main__ import ajcr, a',
            number=10)

import matplotlib.pyplot as plt

fig, axes = plt.subplots(2, 2, figsize=(10, 5))
for i, (name, group) in enumerate(results.stack().groupby(level=0)):
    r, c = i // 2, i % 2
    group.xs(name).plot.barh(ax=axes[r, c], title=name)
fig.tight_layout()

enter image description here

results

enter image description here