从上一次在Pandas DataFrame中发生以来，还剩下几天吗？

Question

假设我有一个Pandas DataFrame df：

Date      Value
01/01/17  0
01/02/17  0
01/03/17  1
01/04/17  0
01/05/17  0
01/06/17  0
01/07/17  1
01/08/17  0
01/09/17  0

对于每一行，我想有效地计算自上次出现以来的天数Value=1。

这样df：

Date      Value    Last_Occurence
01/01/17  0        NaN
01/02/17  0        NaN
01/03/17  1        0
01/04/17  0        1
01/05/17  0        2
01/06/17  0        3
01/07/17  1        0
01/08/17  0        1
01/09/17  0        2

我可以做一个循环：

for i in range(0, len(df)):
    last = np.where(df.loc[0:i,'Value']==1)
    df.loc[i, 'Last_Occurence'] = i-last

但是，对于庞大的数据集而言，效率似乎很低，而且可能还是不正确。

Answer 1

这是NumPy的方法-

def intervaled_cumsum(a, trigger_val=1, start_val = 0, invalid_specifier=-1):
    out = np.ones(a.size,dtype=int)    
    idx = np.flatnonzero(a==trigger_val)
    if len(idx)==0:
        return np.full(a.size,invalid_specifier)
    else:
        out[idx[0]] = -idx[0] + 1
        out[0] = start_val
        out[idx[1:]] = idx[:-1] - idx[1:] + 1
        np.cumsum(out, out=out)
        out[:idx[0]] = invalid_specifier
        return out

很少有关于数组数据的示例来展示涉及触发器和起始值的各种场景的用法：

In [120]: a
Out[120]: array([0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0])

In [121]: p1 = intervaled_cumsum(a, trigger_val=1, start_val=0)
     ...: p2 = intervaled_cumsum(a, trigger_val=1, start_val=1)
     ...: p3 = intervaled_cumsum(a, trigger_val=0, start_val=0)
     ...: p4 = intervaled_cumsum(a, trigger_val=0, start_val=1)
     ...:

In [122]: np.vstack(( a, p1, p2, p3, p4 ))
Out[122]: 
array([[ 0,  1,  1,  1,  0,  0,  1,  0,  0,  1,  1,  1,  1,  1,  0],
       [-1,  0,  0,  0,  1,  2,  0,  1,  2,  0,  0,  0,  0,  0,  1],
       [-1,  1,  1,  1,  2,  3,  1,  2,  3,  1,  1,  1,  1,  1,  2],
       [ 0,  1,  2,  3,  0,  0,  1,  0,  0,  1,  2,  3,  4,  5,  0],
       [ 1,  2,  3,  4,  1,  1,  2,  1,  1,  2,  3,  4,  5,  6,  1]])

用它来解决我们的情况：

df['Last_Occurence'] = intervaled_cumsum(df.Value.values)

样本输出-

In [181]: df
Out[181]: 
       Date  Value  Last_Occurence
0  01/01/17      0              -1
1  01/02/17      0              -1
2  01/03/17      1               0
3  01/04/17      0               1
4  01/05/17      0               2
5  01/06/17      0               3
6  01/07/17      1               0
7  01/08/17      0               1
8  01/09/17      0               2

运行时测试

方法-

# @Scott Boston's soln
def pandas_groupby(df):
    mask = df.Value.cumsum().replace(0,False).astype(bool)
    return df.assign(Last_Occurance=df.groupby(df.Value.astype(bool).\
                                    cumsum()).cumcount().where(mask))

# Proposed in this post
def numpy_based(df):
    df['Last_Occurence'] = intervaled_cumsum(df.Value.values)

时间-

In [33]: df = pd.DataFrame((np.random.rand(10000000)>0.7).astype(int), columns=[['Value']])

In [34]: %timeit pandas_groupby(df)
1 loops, best of 3: 1.06 s per loop

In [35]: %timeit numpy_based(df)
10 loops, best of 3: 103 ms per loop

In [36]: df = pd.DataFrame((np.random.rand(100000000)>0.7).astype(int), columns=[['Value']])

In [37]: %timeit pandas_groupby(df)
1 loops, best of 3: 11.1 s per loop

In [38]: %timeit numpy_based(df)
1 loops, best of 3: 1.03 s per loop