在pandas DataFram的每一行中保留前n个非NaN单元

Question

我有一个Pandas数据帧，每行至少有4个非NaN值， 但位于不同的柱子上：

Index       Col1     Col2      Col3         Col4     Col5  Col6  Col7  Col8 
1991-12-31  100.000 100.000    100.000     89.123   NaN    NaN   NaN   NaN                     
1992-01-31  98.300  101.530    100.000     NaN      92.342 NaN   NaN   NaN                     
1992-02-29  NaN     100.230    98.713      97.602   NaN    NaN   NaN   NaN                     
1992-03-31  NaN     NaN        102.060     93.473   98.123 NaN   NaN   NaN                     
1992-04-30  NaN     102.205    107.755     94.529   94.529 NaN   NaN   NaN

（我只显示前8列）我想把它变成一个每行有4列的数据框。 行应该只包含该日期的前四个非NaN值（从左到右读取）。在

编辑：

每排的顺序很重要。在

Answer 1

方法1:这是一个使用^{}-

pd.DataFrame(justify(df.values, invalid_val=np.nan, axis=1, side='left')[:,:4])

样本运行-

^{pr2}$

方法2:使用特制的面具功能-

def app2(df, N=4):
    a = df.values
    out = np.empty_like(a)
    mask = df.isnull().values
    mask_sorted = np.sort(mask,1)
    out[~mask_sorted] = a[~mask]
    return pd.DataFrame(out[:,:N])

运行时测试保持秩序的工作解决方案-

# Using df from posted question to recreate a bigger one :
df = df.set_index('Index')
df = pd.concat([df] * 10000, ignore_index=1)

In [298]: %timeit app2(df)
100 loops, best of 3: 4.06 ms per loop

In [299]: %timeit pd.DataFrame(justify(df.values, invalid_val=np.nan, axis=1, side='left')[:,:4])
100 loops, best of 3: 4.78 ms per loop

In [300]: %timeit df.apply(sorted, key=np.isnan, axis=1).iloc[:, :4]
1 loop, best of 3: 4.05 s per loop

Answer 2

如果顺序不重要，可以沿着第一个轴调用np.sort。在

df = df.set_index('Index')   # ignore if `Index` already is the index

pd.DataFrame(np.sort(df.values, axis=1)[:, :4], 
           columns=np.arange(1, 5)).add_prefix('Col')

     Col1     Col2     Col3     Col4
0  89.123  100.000  100.000  100.000
1  92.342   98.300  100.000  101.530
2  97.602   98.713  100.230      NaN
3  93.473   98.123  102.060      NaN
4  94.529   94.529  102.205  107.755

这比我的第二个解决方案快得多，所以如果可能的话，一定要考虑这个。在

---

如果顺序很重要，请调用sorted+apply，并获取结果的前4列。在

^{pr2}$

---

计时
以下是我的答案的时间安排-

df = pd.concat([df] * 10000, ignore_index=1)

%timeit df.apply(sorted, key=np.isnan, axis=1).iloc[:, :4]
1 loop, best of 3: 8.45 s per loop

pd.DataFrame(np.sort(df.values, axis=1)[:, :4], 
           columns=np.arange(1, 5)).add_prefix('Col')    
100 loops, best of 3: 4.76 ms per loop

Answer 3

您可以使用：

#if necessary
#df = df.set_index('Index')

df = df.apply(lambda x: pd.Series(x.dropna().values), axis=1).iloc[:, :4]
print (df)
                  0        1        2       3
Index                                        
1991-12-31  100.000  100.000  100.000  89.123
1992-01-31   98.300  101.530  100.000  92.342
1992-02-29  100.230   98.713   97.602     NaN
1992-03-31  102.060   93.473   98.123     NaN
1992-04-30  102.205  107.755   94.529  94.529

或者为了获得更好的性能，请使用numpy-处理需求，每行至少有4个非值：

^{pr2}$

计时：

        Index   Col1     Col2     Col3    Col4    Col5  Col6  Col7  Col8
0  1991-12-31  100.0  100.000  100.000  89.123     NaN   NaN   NaN   NaN
1  1992-01-31   98.3  101.530  100.000     NaN  92.342   NaN   NaN   NaN
2  1992-02-29    NaN  100.230   98.713  97.602     NaN   NaN   NaN   1.0
3  1992-03-31    NaN      NaN  102.060  93.473  98.123   NaN   NaN   1.0
4  1992-04-30    NaN  102.205  107.755  94.529  94.529   NaN   NaN   NaN

df = df.set_index('Index')

df = pd.concat([df] * 10000, ignore_index=1)

In [260]: %timeit pd.DataFrame(justify(df.values, invalid_val=np.nan, axis=1, side='left')[:,:4])
100 loops, best of 3: 6.78 ms per loop

In [261]: %%timeit a = df.values
     ...: pd.DataFrame(a[~np.isnan(a)].reshape(a.shape[0],-1)[:, :4], index=df.index)
     ...: 
100 loops, best of 3: 2.11 ms per loop

In [262]: %timeit pd.DataFrame(np.sort(df.values, axis=1)[:, :4], columns=np.arange(1, 5)).add_prefix('Col')
100 loops, best of 3: 5.28 ms per loop

In [263]: %timeit pd.DataFrame(mask_app(df.values)[:,:4])
100 loops, best of 3: 8.68 ms per loop