Python 对数据帧中的连续数据进行计数，并在出现这种情况时获取索引

我有一个带有整数列名的

pandas.DataFrame

，它有0和1。输入的一个示例：

    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/和大于等于2的所有连续的索引，遍历列，并返回出现连续索引数组的索引（开始、结束）

首选的输出是3D数据帧，其中子列“count”和“index”表示输入中的整数列名

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)     

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我想应该用

itertools.groupby

解决这个问题，但是仍然无法找出如何将它应用到这样的问题，在这个问题中，

groupby

结果及其索引都被提取出来。

这里有一种计算所需运行长度的方法：

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]

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)

{
 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))], 
}

测试代码：

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]

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)

{
 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))], 
}

结果：

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]

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)

{
 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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相关但不完全相同：这是一个非常聪明的解决方案！谢谢！