Pandas Groupby 应用函数非常慢，循环每个组 > 应用函数 > 添加结果作为新列

Question

我有库存数据，我试图在其中查找价格是否连续 5 天下降。示例数据：

我按 Symbol 分组。

df_s = df.groupby(['Symbol'])

我想为每个组申请的功能是：

def strictly_decreasing(L):
    #---it will return True if L(list) is decreasing, eg:[7,5,3,2] = True---
    return all(x>y for x, y in zip(L, L[1:]))

def strick_dec(group):
    dec = []
    for i in range(0,len(group)-4):
        chec = group["close"][i:i+5]
        dec.append(strictly_decreasing(list(chec)))
    dummy = [False,False,False,False]                # because we skiped last 4 values 
    final =  dec + dummy                             # can also be used as dummy + dec, up to you
    group['Strictly_decreasing'] = final             # adding the results as new column
    return group

df_new = df_s.apply(strick_dec)

我们怎样才能加快这个过程？太浪费时间了。

Answer 1

它可能永远不会很快，但使用 pandas 方法只会加快一点点。嵌套循环并不是真正需要的。尝试这样的事情：

import datetime
import pandas
import random
import itertools

# Create some test data
now = datetime.datetime.now()
df = pandas.DataFrame(
    itertools.chain.from_iterable(
        [
            [
                {
                    "symbol": "".join(symbol),
                    "date": now + pandas.Timedelta(-i, unit="D"),
                    "close": random.randint(10, 100) + random.random(),
                    "volume": random.randint(20000, 1000000),
                }
                for i in range(60)
            ]
            for symbol in itertools.combinations("ABCDEFGHIJKLM", 4)
        ]
    )
)


def check_decreasing(group: pandas.DataFrame, column: str = "close") -> pandas.DataFrame:

    # Add shifted columns to show the previous value of close in the next column
    for i in range(4, 0, -1):
        group[f"{column}_minus_{i}"] = group[f"{column}"].shift(i)

    # Use pandas.is_monotonic_decrease to check if the values are decreasing
    group["is_monotonic_decreasing"] = group[[f"{column}_minus_{i}" for i in range(4, 0, -1)] + [f"{column}"]].apply(lambda row: row.is_monotonic_decreasing, axis=1)

    # Remove the shifted columns (no longer needed)
    group = group.drop(columns=[f"{column}_minus_{i}" for i in range(4, 0, -1)])

    # Return the group
    return group


# Fix some rows for testing (random will not always give results), this will create artificial monotonic decrease in the first 10 rows
for i in range(10):
    df.at[i, "close"] = 100 - i*5

# Apply the function
df = df.groupby("symbol").apply(check_decreasing, column="close")

输出：

      symbol                       date       close  volume  \
0       ABCD 2020-11-30 09:00:16.102408  100.000000  631890   
1       ABCD 2020-11-29 09:00:16.102408   95.000000  717153   
2       ABCD 2020-11-28 09:00:16.102408   90.000000  248423   
3       ABCD 2020-11-27 09:00:16.102408   85.000000  987648   
4       ABCD 2020-11-26 09:00:16.102408   80.000000  613279   
...      ...                        ...         ...     ...   
42895   JKLM 2020-10-06 09:00:16.102408   31.103065  740687   
42896   JKLM 2020-10-05 09:00:16.102408   75.330438  794853   
42897   JKLM 2020-10-04 09:00:16.102408   47.115309  279714   
42898   JKLM 2020-10-03 09:00:16.102408   15.527207  972621   
42899   JKLM 2020-10-02 09:00:16.102408   60.094327  765083   

       is_monotonic_decreasing  
0                        False  
1                        False  
2                        False  
3                        False  
4                         True  
...                        ...  
42895                    False  
42896                    False  
42897                    False  
42898                    False  
42899                    False  

[42900 rows x 5 columns]

Pandas Groupby 应用函数非常慢，循环每个组 > 应用函数 > 添加结果作为新列

Pandas Groupby apply function is very slow , Looping every group > applying function>adding results as new column

python

pandas

pandas-apply

pandas-groupby