计算数据帧中的连续数，并在发生这种情况的情况下获取索引 [英] Count consecutive ones in a dataframe and get indices where this occurs

问题描述

我有一个pandas.DataFrame，带有整数列名，该列名有零和一.输入的示例:

I have a pandas.DataFrame with integer column names, which has zeroes and ones. An example of the input:

    12  13  14  15
1   0   0   1   0
2   0   0   1   1
3   1   0   0   1
4   1   1   0   1
5   1   1   1   0
6   0   0   1   0
7   0   0   1   1
8   1   1   0   1
9   0   0   1   1
10  0   0   1   1
11  1   1   0   1
12  1   1   1   1
13  1   1   1   1
14  1   0   1   1
15  0   0   1   1

我需要计算长度/总和为> = 2的所有连续数，遍历各列，并返回出现连续数数组(起始，结束)的索引.

I need to count all consecutive ones which has a length/sum which is >=2, iterating through columns and returning also indices where an array of the consecutive ones occurs (start, end).

首选输出将是3D DataFrame，其中子列"count"和"indices"是指输入中的整数列名称.

The preferred output would be a 3D DataFrame, where subcolumns "count" and "indices" refer to integer column names from the input.

示例输出如下所示:

12              13              14              15
count   indices count   indices count   indices count   indices
    3     (3,5)     2     (4,5)     2     (1,2)     3     (2,4)
    4   (11,14)     3   (11,13)     3     (5,7)     9    (7,15)
                                    2    (9,10) 
                                    4   (12,15)     

我想应该用itertools.groupby来解决它，但是仍然无法弄清楚如何将其应用到这样的问题中，在该问题中，同时提取了groupby的结果及其索引.

I suppose it should be solved with itertools.groupby, but still can't figure out how to apply it to such problem, where both groupby results and its indices are being extracted.

推荐答案

这里是计算所需游程长度的一种方法:

Here is one way to calculate the desired run lengths:

代码:

def min_run_length(series):
    terminal = pd.Series([0])
    diffs = pd.concat([terminal, series, terminal]).diff()
    starts = np.where(diffs == 1)
    ends = np.where(diffs == -1)
    return [(e-s, (s, e-1)) for s, e in zip(starts[0], ends[0])
            if e - s >= 2]

测试代码:

df = pd.read_fwf(StringIO(u"""
    12  13  14  15
    0   0   1   0
    0   0   1   1
    1   0   0   1
    1   1   0   1
    1   1   1   0
    0   0   1   0
    0   0   1   1
    1   1   0   1
    0   0   1   1
    0   0   1   1
    1   1   0   1
    1   1   1   1
    1   1   1   1
    1   0   1   1
    0   0   1   1"""), header=1)
print(df.dtypes)

indices = {cname: min_run_length(df[cname]) for cname in df.columns}
print(indices)

结果:

{
 u'12': [(3, (3, 5)), (4, (11, 14))], 
 u'13': [(2, (4, 5)), (3, (11, 13))], 
 u'14': [(2, (1, 2)), (3, (5, 7)), (2, (9, 10)), (4, (12, 15))]
 u'15': [(3, (2, 4)), (9, (7, 15))], 
}

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