如何找到熊猫时间序列的最后一个局部最大值？

Question

给定以下数据框：

pd.DataFrame([['a', 1], ['b', 3], ['c', 7], ['d', 4], ['e', 1], ['f', 2], ['g', 9], ['h', 4], ['i', 0]])

有没有更好的方法来找到最后一个局部最大值（'g'），而不是反向逐行迭代并搜索大于当前行的前一行？

这是我目前正在使用的，必须有更有效的东西：

df.columns = ['x', 'y']

first = True
prev_val = None
prev_row = None

for index, row in df[::-1].iterrows():
    if first:
        prev_val = row['y']
        prev_row = row['x']
        first = False
    else:    
        if row['y'] >= prev_val:
            prev_val = row['y']
            prev_row = row['x']
        else:
            break

Answer 1

使用scipy.signal.argrelextrema:

from scipy.signal import argrelextrema

a = df.iat[argrelextrema(df[1].values, np.greater)[0][-1], 0]
print (a)
g

Answer 2

一般而言，您应该尽量避免手动循环，尤其是使用iterrows。一个例外是numba，它在较低级别高效地执行迭代：

from numba import jit

df = pd.DataFrame([['a', 1], ['b', 3], ['c', 9], ['d', 4], ['e', 1],
                   ['f', 2], ['g', 7], ['h', 4], ['i', 0]])

@jit(nopython=True)
def local_max_idx(A):
    for i in range(1, len(A)):
        if A[-(i+1)] < A[-i]:
            return -i

res = df[0].iat[local_max_idx(df[1].values)]  # 'g'

性能基准测试

n = 1000000
df = pd.Series([0] + list(range(n, 0, -1))).to_frame().reset_index()
df.columns = [0, 1]

@jit(nopython=True)
def local_max_idx(A):
    for i in range(1, len(A)):
        if A[-(i+1)] < A[-i]:
            return -i

from scipy.signal import argrelextrema

%timeit df.iat[argrelextrema(df[1].values, np.greater)[0][-1], 0]  # 46.1 ms per loop
%timeit df[0].iat[local_max_idx(df[1].values)]                     # 1.59 ms per loop