3

我有一个像这样的数据框

                       OPEN    HIGH     LOW   CLOSE         VOL
2012-01-01 19:00:00  449000  449000  449000  449000  1336303000
2012-01-01 20:00:00     NaN     NaN     NaN     NaN         NaN
2012-01-01 21:00:00     NaN     NaN     NaN     NaN         NaN
2012-01-01 22:00:00     NaN     NaN     NaN     NaN         NaN
2012-01-01 23:00:00     NaN     NaN     NaN     NaN         NaN
...
                         OPEN      HIGH       LOW     CLOSE          VOL
2013-04-24 14:00:00  11700000  12000000  11600000  12000000  20647095439
2013-04-24 15:00:00  12000000  12399000  11979000  12399000  23997107870
2013-04-24 16:00:00  12399000  12400000  11865000  12100000   9379191474
2013-04-24 17:00:00  12300000  12397995  11850000  11850000   4281521826
2013-04-24 18:00:00  11850000  11850000  10903000  11800000  15546034128

我需要NaN按照这个规则填写

当 OPEN、HIGH、LOW、CLOSE 为 NaN 时，

  • 将 VOL 设置为 0
  • 将 OPEN、HIGH、LOW、CLOSE 设置为之前的 CLOSE 蜡烛值

否则保留 NaN

4

3 回答 3

1

由于其他两个答案都不起作用，因此这是一个完整的答案。

我在这里测试两种方法。第一个基于 working4coin 对 hd1 答案的评论，第二个是较慢的纯 python 实现。很明显，python 实现应该更慢，但我决定对这两种方法进行计时，以确保并量化结果。

def nans_to_prev_close_method1(data_frame):
    data_frame['volume'] = data_frame['volume'].fillna(0.0)  # volume should always be 0 (if there were no trades in this interval)
    data_frame['close'] = data_frame.fillna(method='pad')  # ie pull the last close into this close
    # now copy the close that was pulled down from the last timestep into this row, across into o/h/l
    data_frame['open'] = data_frame['open'].fillna(data_frame['close']) 
    data_frame['low'] = data_frame['low'].fillna(data_frame['close'])
    data_frame['high'] = data_frame['high'].fillna(data_frame['close'])

方法 1 完成了 c 中的大部分繁重工作（在 pandas 代码中），因此应该很快。

缓慢的python方法（方法2）如下所示

def nans_to_prev_close_method2(data_frame):
    prev_row = None
    for index, row in data_frame.iterrows():
        if np.isnan(row['open']):  # row.isnull().any():
            pclose = prev_row['close']
            # assumes first row has no nulls!!
            row['open'] = pclose
            row['high'] = pclose
            row['low'] = pclose
            row['close'] = pclose
            row['volume'] = 0.0
        prev_row = row

测试他们两个的时间：

df = trades_to_ohlcv(PATH_TO_RAW_TRADES_CSV, '1s') # splits raw trades into secondly candles
df2 = df.copy()

wrapped1 = wrapper(nans_to_prev_close_method1, df)
wrapped2 = wrapper(nans_to_prev_close_method2, df2)

print("method 1: %.2f sec" % timeit.timeit(wrapped1, number=1))
print("method 2: %.2f sec" % timeit.timeit(wrapped2, number=1))

结果是：

method 1:   0.46 sec
method 2: 151.82 sec

显然方法 1 要快得多（大约快 330 倍）。

于 2017-12-28T23:30:14.777 回答
0

这说明了pandas缺失数据的行为。您正在寻找的咒语是 fillna 方法，它接受一个值：

In [1381]: df2
Out[1381]: 
        one       two     three four   five           timestamp
a       NaN  1.138469 -2.400634  bar   True                 NaT
c       NaN  0.025653 -1.386071  bar  False                 NaT
e  0.863937  0.252462  1.500571  bar   True 2012-01-01 00:00:00
f  1.053202 -2.338595 -0.374279  bar   True 2012-01-01 00:00:00
h       NaN -1.157886 -0.551865  bar  False                 NaT

In [1382]: df2.fillna(0)
Out[1382]: 
        one       two     three four   five           timestamp
a  0.000000  1.138469 -2.400634  bar   True 1970-01-01 00:00:00
c  0.000000  0.025653 -1.386071  bar  False 1970-01-01 00:00:00
e  0.863937  0.252462  1.500571  bar   True 2012-01-01 00:00:00
f  1.053202 -2.338595 -0.374279  bar   True 2012-01-01 00:00:00
h  0.000000 -1.157886 -0.551865  bar  False 1970-01-01 00:00:00

你甚至可以向前和向后传播它们：

In [1384]: df
Out[1384]: 
        one       two     three
a       NaN  1.138469 -2.400634
c       NaN  0.025653 -1.386071
e  0.863937  0.252462  1.500571
f  1.053202 -2.338595 -0.374279
h       NaN -1.157886 -0.551865

In [1385]: df.fillna(method='pad')
Out[1385]: 
        one       two     three
a       NaN  1.138469 -2.400634
c       NaN  0.025653 -1.386071
e  0.863937  0.252462  1.500571
f  1.053202 -2.338595 -0.374279
h  1.053202 -1.157886 -0.551865

对于您的具体情况，我认为您需要执行以下操作：

df['VOL'].fillna(0)
df.fillna(df['CLOSE'])
于 2013-05-09T16:41:50.840 回答
0

这是通过掩蔽的方法

模拟带有一些孔的框架（A 是您的“关闭”字段）

In [20]: df = DataFrame(randn(10,3),index=date_range('20130101',periods=10,freq='min'),
            columns=list('ABC'))

In [21]: df.iloc[1:3,:] = np.nan

In [22]: df.iloc[5:8,1:3] = np.nan

In [23]: df
Out[23]: 
                            A         B         C
2013-01-01 00:00:00 -0.486149  0.156894 -0.272362
2013-01-01 00:01:00       NaN       NaN       NaN
2013-01-01 00:02:00       NaN       NaN       NaN
2013-01-01 00:03:00  1.788240 -0.593195  0.059606
2013-01-01 00:04:00  1.097781  0.835491 -0.855468
2013-01-01 00:05:00  0.753991       NaN       NaN
2013-01-01 00:06:00 -0.456790       NaN       NaN
2013-01-01 00:07:00 -0.479704       NaN       NaN
2013-01-01 00:08:00  1.332830  1.276571 -0.480007
2013-01-01 00:09:00 -0.759806 -0.815984  2.699401

我们都是南

In [24]: mask_0 = pd.isnull(df).all(axis=1)

In [25]: mask_0
Out[25]: 
2013-01-01 00:00:00    False
2013-01-01 00:01:00     True
2013-01-01 00:02:00     True
2013-01-01 00:03:00    False
2013-01-01 00:04:00    False
2013-01-01 00:05:00    False
2013-01-01 00:06:00    False
2013-01-01 00:07:00    False
2013-01-01 00:08:00    False
2013-01-01 00:09:00    False
Freq: T, dtype: bool

我们要传播的那些 A

In [26]: mask_fill = pd.isnull(df['B']) & pd.isnull(df['C'])

In [27]: mask_fill
Out[27]: 
2013-01-01 00:00:00    False
2013-01-01 00:01:00     True
2013-01-01 00:02:00     True
2013-01-01 00:03:00    False
2013-01-01 00:04:00    False
2013-01-01 00:05:00     True
2013-01-01 00:06:00     True
2013-01-01 00:07:00     True
2013-01-01 00:08:00    False
2013-01-01 00:09:00    False
Freq: T, dtype: bool

先发扬光大

In [28]: df.loc[mask_fill,'C'] = df['A']

In [29]: df.loc[mask_fill,'B'] = df['A']

填充 0

In [30]: df.loc[mask_0] = 0

完毕

In [31]: df
Out[31]: 
                            A         B         C
2013-01-01 00:00:00 -0.486149  0.156894 -0.272362
2013-01-01 00:01:00  0.000000  0.000000  0.000000
2013-01-01 00:02:00  0.000000  0.000000  0.000000
2013-01-01 00:03:00  1.788240 -0.593195  0.059606
2013-01-01 00:04:00  1.097781  0.835491 -0.855468
2013-01-01 00:05:00  0.753991  0.753991  0.753991
2013-01-01 00:06:00 -0.456790 -0.456790 -0.456790
2013-01-01 00:07:00 -0.479704 -0.479704 -0.479704
2013-01-01 00:08:00  1.332830  1.276571 -0.480007
2013-01-01 00:09:00 -0.759806 -0.815984  2.699401
于 2013-05-09T17:14:42.130 回答