Question

我想将连续的NaN值合并成片。使用numpy或pandas可以做到这一点吗？

l = [
    (996, np.nan), (997, np.nan), (998, np.nan),
    (999, -47.3), (1000, -72.5), (1100, -97.7),
    (1200, np.nan), (1201, np.nan), (1205, -97.8),
    (1300, np.nan), (1302, np.nan), (1305, -97.9),
    (1400, np.nan), (1405, -97.10), (1408, np.nan)
]
l = pd.Series(dict(l))

预期结果：

[
    (slice(996, 999, None), array([nan, nan, nan])),
    (999, -47.3),
    (1000, -72.5),
    (1100, -97.7),
    (slice(1200, 1202, None), array([nan, nan])),
    (1205, -97.8),
    (slice(1300, 1301, None), array([nan])),
    (slice(1302, 1303, None), array([nan])),
    (1305, -97.9),
    (slice(1400, 1401, None), array([nan])),
    (1405, -97.1),
    (slice(1408, 1409, None), array([nan]))
]

具有二维的numpy数组也可以，而不是元组列表

更新2019/05/31 ：我刚刚意识到，如果我只使用字典而不是熊猫系列，算法效率会更高

Answer 1

您要的是全角或极角情况，南等式，每对中的第一个元素是切片或单个值，第二个元素是np.array或单个值。

对于如此复杂的要求，我将仅依靠普通的Python非矢量化方式：

def trans(ser):
    def build(last, cur, val):
        if cur == last + 1:
            if np.isnan(val):
                return (slice(last, cur), np.array([np.nan]))
            else:
                return (last, val)
        else:
            return (slice(last, cur), np.array([val] * (cur - last)))
    last = ser.iloc[0]
    old = last_index = ser.index[0]
    resul = []
    for i in ser.index[1:]:
        val = ser[i]
        if ((val != last) and not(np.isnan(val) and np.isnan(last))) \
           or i != old + 1:
            resul.append(build(last_index, old + 1, last))
            last_index = i
            last = val
        old = i
    resul.append(build(last_index, old+1, last))
    return resul

它提供了接近预期结果的东西：

[(slice(996, 999, None), array([nan, nan, nan])),
 (999, -47.3),
 (1000, -72.5),
 (1100, -97.7),
 (slice(1200, 1202, None), array([nan, nan])),
 (1205, -97.8),
 (slice(1300, 1301, None), array([nan])),
 (slice(1302, 1303, None), array([nan])),
 (1305, -97.9),
 (slice(1400, 1401, None), array([nan])),
 (1405, -97.1),
 (slice(1408, 1409, None), array([nan]))]

Answer 2

按cumsum的{{1}}分组是个好主意，但是我们需要过滤掉每个子系列中的第一个非空值，因此我们可以按notnull对分组：

(cumsum, notnull)

出局：

# convert series to frame, 
# don't know why series only doesn't work
df = l.to_frame(name='val')

df['notnull'] = df['val'].notnull()
g = df.groupby([ df['notnull'].cumsum(), 'notnull']).val

[(v.index, v.values) for i, v in g]

编辑：考虑连续索引并更新切片：

[(Int64Index([996, 997, 998], dtype='int64'), array([nan, nan, nan])),
 (Int64Index([1200, 1201], dtype='int64'), array([nan, nan])),
 (Int64Index([1300, 1302, 1400, 1402], dtype='int64'),
  array([nan, nan, nan, nan])),
 (Int64Index([999], dtype='int64'), array([-47.3])),
 (Int64Index([1000], dtype='int64'), array([-72.5])),
 (Int64Index([1100], dtype='int64'), array([-97.7])),
 (Int64Index([1202], dtype='int64'), array([-97.1]))]

给予：

# convert group to slices
def get_slice(x):
    idx_min, idx_max = x.index.min(), x.index.max()

    if len(x) >1:
        return (slice(idx_min, idx_max+1), x.values)
    elif x.isna().any():
        return (slice(idx_min, idx_min+1), x.values)
    else:
        return (idx_min, x[idx_min])

df['notnull'] = df['val'].notnull()

# non-continuous indices
df['sep'] = (df.index != df.index.to_series().shift() + 1).cumsum()

g = df.groupby(['sep', df['notnull'].cumsum(), 'notnull']).val

g.apply(get_slice).values.tolist()

如何在一组切片中将来自熊猫系列的连续NaN值分组？

2 个答案: