熊猫索引:识别相同值连续重复的子范围

原始问题

哪里 idx = pd.Index(['X', 'C', 'C', 'C', 'Q', 'Q', 'Q', 'Q', 'A', 'P', 'P']) .

有点不传统,但应该有效,而且似乎速度要快得多:

# Get a new Index which is the unique duplicated values in `idx`
un = idx[idx.duplicated(keep=False)].unique()

# Call `get_loc` on `idx` for each member of `un` above  
# `np.where` gets position of True in boolean Index
res = []
for i in un:
    w = np.where(idx.get_loc(i))[0]
    # w[0], w[-1] analogous to v.min(), v.max() from @MaxU's answer
    res.append((w[0], w[-1], i))

print(res)
# [(1, 3, 'C'), (4, 7, 'Q'), (9, 10, 'P')]

%timeit myanswer()
105 µs ± 3.19 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

%timeit maxu()
1.21 ms ± 116 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

未注释:

un = idx[idx.duplicated(keep=False)].unique()
res = []
for i in un:
    w = np.where(idx.get_loc(i))[0]
    res.append((w[0], w[-1], i))

已编辑的问题

哪里 idx = pd.Index(['X', 'C', 'C', 'C', 'Q', 'Q', 'Q', 'Q', 'X', 'P', 'P']) .

到达 un 这里,首先得到一个布尔索引,当一个值等于它之前或之后的值时,该索引为True,否则为False。这类似于 idx.duplicated(keep=False) 第一部分。

b = (Series(idx).shift() == idx) | (Series(idx).shift(-1) == idx)
un = idx[b].unique()
# Rest should be the same

这里有一种方法:

In [107]: ix = pd.Series(idx.values)

In [108]: [(v.min(), v.max(),k) for k,v in ix.groupby(ix).groups.items() if len(v) > 1]
Out[108]: [(1, 3, 'C'), (9, 10, 'P'), (4, 7, 'Q')]