如何计算DataFrame中连续TRUE的数量? [英] How can I count the number of consecutive TRUEs in a DataFrame?

问题描述

我有一个由True和False组成的数据集.

I have a dataset made of True and False.

Sample Table:
       A      B      C
0  False   True  False
1  False  False  False
2   True   True  False
3   True   True   True
4  False   True  False
5   True   True   True
6   True  False  False
7   True  False   True
8  False   True   True
9   True  False  False

我想计算每一列的连续True值的数量，如果连续的True系列超过一个，我想获取最大值.

I want to count the number of consecutive True values for every column, and if there's more than one consecutive True series, I want to get the max of it.

对于上表，我会得到:

length = [3, 4, 2]

我发现了类似的线程，但没有一个解决我的问题.

I found similar threads but none resolved my problem.

自从我这样做以后，将有更多的列(产品)，因此无论列名如何，我都需要对整个表执行此操作，并得到一个数组作为结果.

Since I do and will have many more columns(products), I need to do this regardless of the column name, for the whole table and get an array as the result.

如果可能的话，我想学习最长序列的第一个true的索引，也就是这个最长的true系列开始的地方，所以结果就是这个:

And if possible, I'd like to learn the index of the first true of the longest sequence aka where this longest true series starts, so the result would be for this one:

index = [5, 2, 7]

推荐答案

我们基本上会利用两种理念- Catching shifts on compared array 和 Offsetting each column results so that we could vectorize it .

We would basically leverage two philosophies - Catching shifts on compared array and Offsetting each column results so that we could vectorize it.

因此，在设置了此意图之后，这是一种获得预期结果的方法-

So, with that intention set, here's one way to achieve the desired results -

def maxisland_start_len_mask(a, fillna_index = -1, fillna_len = 0):
    # a is a boolean array

    pad = np.zeros(a.shape[1],dtype=bool)
    mask = np.vstack((pad, a, pad))

    mask_step = mask[1:] != mask[:-1]
    idx = np.flatnonzero(mask_step.T)
    island_starts = idx[::2]
    island_lens = idx[1::2] - idx[::2]
    n_islands_percol = mask_step.sum(0)//2

    bins = np.repeat(np.arange(a.shape[1]),n_islands_percol)
    scale = island_lens.max()+1

    scaled_idx = np.argsort(scale*bins + island_lens)
    grp_shift_idx = np.r_[0,n_islands_percol.cumsum()]
    max_island_starts = island_starts[scaled_idx[grp_shift_idx[1:]-1]]

    max_island_percol_start = max_island_starts%(a.shape[0]+1)

    valid = n_islands_percol!=0
    cut_idx = grp_shift_idx[:-1][valid]
    max_island_percol_len = np.maximum.reduceat(island_lens, cut_idx)

    out_len = np.full(a.shape[1], fillna_len, dtype=int)
    out_len[valid] = max_island_percol_len
    out_index = np.where(valid,max_island_percol_start,fillna_index)
    return out_index, out_len

样品运行-

# Generic case to handle all 0s columns
In [112]: a
Out[112]: 
array([[False, False, False],
       [False, False, False],
       [ True, False, False],
       [ True, False,  True],
       [False, False, False],
       [ True, False,  True],
       [ True, False, False],
       [ True, False,  True],
       [False, False,  True],
       [ True, False, False]])

In [117]: starts,lens = maxisland_start_len_mask(a, fillna_index=-1, fillna_len=0)

In [118]: starts
Out[118]: array([ 5, -1,  7])

In [119]: lens
Out[119]: array([3, 0, 2])

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