Question

我正在尝试用零填充数据框，但我不想触及领先的NaN ：

rng = pd.date_range('2016-06-01', periods=9, freq='D')
df = pd.DataFrame({'data': pd.Series([np.nan]*3 + [20, 30, 40] + [np.nan]*3, rng)})

2016-06-01     NaN
2016-06-02     NaN
2016-06-03     NaN
2016-06-04    20.0
2016-06-05    30.0
2016-06-06    40.0
2016-06-07     NaN
2016-06-08     NaN
2016-06-09     NaN

填充/替换后我想要的是：

pd.DataFrame({'data': pd.Series([np.nan]*3 + [20, 30, 40] + [0.]*3, rng)})

2016-06-01     NaN
2016-06-02     NaN
2016-06-03     NaN
2016-06-04    20.0
2016-06-05    30.0
2016-06-06    40.0
2016-06-07     0.0
2016-06-08     0.0
2016-06-09     0.0

由于fillna()仅允许值或方法而fillna(0)替换所有NaN，包括前导，所以我希望替换可以跳到此处，但

df.replace([np.nan], 0, method='ffill')

也取代所有NaN。

如何仅在第一个非NaN值之后填充值，同时还有多个数据列？

Answer 1

您可以使用last_valid_index()功能

来完成

In [80]: df
Out[80]:
            data  data1  data2
2016-06-01   NaN    NaN    NaN
2016-06-02   NaN    NaN   10.0
2016-06-03   NaN   20.0   20.0
2016-06-04  20.0   30.0   20.0
2016-06-05   NaN   40.0    NaN
2016-06-06  40.0   30.0   40.0
2016-06-07   NaN    NaN    NaN
2016-06-08   NaN    NaN    NaN
2016-06-09   NaN    NaN    NaN

In [81]: %paste
first_valid_idx = df.apply(lambda x: x.first_valid_index()).to_frame()
df = df.fillna(0)
for ix, r in first_valid_idx.iterrows():
    df.loc[df.index < r[0], ix] = np.nan
## -- End pasted text --

In [82]: df
Out[82]:
            data  data1  data2
2016-06-01   NaN    NaN    NaN
2016-06-02   NaN    NaN   10.0
2016-06-03   NaN   20.0   20.0
2016-06-04  20.0   30.0   20.0
2016-06-05   0.0   40.0    0.0
2016-06-06  40.0   30.0   40.0
2016-06-07   0.0    0.0    0.0
2016-06-08   0.0    0.0    0.0
2016-06-09   0.0    0.0    0.0

In [83]: first_valid_idx
Out[83]:
               0
data  2016-06-04
data1 2016-06-03
data2 2016-06-02

OLD回答：

In [38]: df.loc[df.index > df.data.last_valid_index(), 'data'] = 0

In [39]: df
Out[39]:
            data
2016-06-01   NaN
2016-06-02   NaN
2016-06-03   NaN
2016-06-04  20.0
2016-06-05  30.0
2016-06-06  40.0
2016-06-07   0.0
2016-06-08   0.0
2016-06-09   0.0

Answer 2

我认为您可以group NaN的{{1}} print (df.data.notnull().cumsum()) 2016-06-01 0 2016-06-02 0 2016-06-03 0 2016-06-04 1 2016-06-05 2 2016-06-06 3 2016-06-07 3 2016-06-08 3 2016-06-09 3 Freq: D, Name: data, dtype: int32 print (df.data.mask(df.data.notnull().cumsum() != 0, df.data.fillna(0))) 2016-06-01 NaN 2016-06-02 NaN 2016-06-03 NaN 2016-06-04 20.0 2016-06-05 30.0 2016-06-06 40.0 2016-06-07 0.0 2016-06-08 0.0 2016-06-09 0.0 Freq: D, Name: data, dtype: float64 isnull，然后cumsum找到fillna所有其他值：

df = pd.DataFrame({'data': pd.Series([np.nan]*3 + [20, 30, 40] + [np.nan]*3, rng), 
                   'data1': pd.Series([np.nan]*2 + [20, 30, 40,30] + [np.nan]*3, rng),
                   'data2': pd.Series([np.nan]*1 + [10,20, 20, 30, 40] + [np.nan]*3, rng)})

print (df.mask(df.notnull().cumsum() != 0, df.fillna(0)))
            data  data1  data2
2016-06-01   NaN    NaN    NaN
2016-06-02   NaN    NaN   10.0
2016-06-03   NaN   20.0   20.0
2016-06-04  20.0   30.0   20.0
2016-06-05  30.0   40.0   30.0
2016-06-06  40.0   30.0   40.0
2016-06-07   0.0    0.0    0.0
2016-06-08   0.0    0.0    0.0
2016-06-09   0.0    0.0    0.0

编辑：

对于多列，它也很好用：

print (df.mask(df.notnull().cummax(), df.fillna(0)))
            data  data1  data2
2016-06-01   NaN    NaN    NaN
2016-06-02   NaN    NaN   10.0
2016-06-03   NaN   20.0   20.0
2016-06-04  20.0   30.0   20.0
2016-06-05  30.0   40.0   30.0
2016-06-06  40.0   30.0   40.0
2016-06-07   0.0    0.0    0.0
2016-06-08   0.0    0.0    0.0
2016-06-09   0.0    0.0    0.0

EDIT2 DSM评论 - 更好的是使用cummax：

$json_object=json_decode($jsonString);

零填充数据帧前向填充

2 个答案: