python-2.7 – 数据帧中的Pandas列表理解

如果满足某些条件,我想在当前存储在(第1行)的第二天打开价格并将其存储在新列中.

df['b']=''

df['shift']=''

df['shift']=df['open'].shift(-1)

df['b']=df[x for x in df['shift'] if df["MA10"]>df["MA100"]]

解决方法

>>> df = pd.read_csv("bondstack.csv")
>>> df["shift"] = df["open"].shift(-1)
>>> df["b"] = df.apply(lambda row: row["shift"] if row["MA10"] > row["MA100"] else np.nan,axis=1)

哪个产生

>>> df[["MA10","MA100","shift","b"]][:10]
        MA10      MA100      shift          b
0  16.915625  17.405625  16.734375        NaN
1  16.871875  17.358750  17.171875        NaN
2  16.893750  17.317187  17.359375        NaN
3  16.950000  17.279062  17.359375        NaN
4  17.137500  17.254062  18.640625        NaN
5  17.365625  17.229063  18.921875  18.921875
6  17.550000  17.200312  18.296875  18.296875
7  17.681250  17.177500  18.640625  18.640625
8  17.812500  17.159375  18.609375  18.609375
9  17.943750  17.142813  18.234375  18.234375

对于更加矢量化的方法,您可以使用

>>> df = pd.read_csv("bondstack.csv")
>>> df["b"] = np.nan
>>> df["b"][df["MA10"] > df["MA100"]] = df["open"].shift(-1)

或者我的首选方法：

>>> df = pd.read_csv("bondstack.csv")
>>> df["b"] = df["open"].shift(-1).where(df["MA10"] > df["MA100"])

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