如何编写这个 for 循环以在 Python 中更快地处理

运行一系列不同的示例，下面代码中的第二个方法对于示例数据集来说大约快了X75:

import pandas as pd, numpy as np
from random import randint
import time

data = [randint(90, 120) for i in range(10000)]

df1 = pd.DataFrame({'price': data})

t0 = time.time()
df1['updown'] = np.nan
count = df1.shape[0]
for i in range(count):
    j = 1
    up = df1.price.iloc[i] + 2
    down = up - 4
    while (pos := i + j) < count:
        if(value := df1.price.iloc[pos]) >= up:
            df1.loc[i, 'updown'] = "Up"
            break
        elif value <= down:
            df1.loc[i, 'updown'] = "Down"
            break
        else:
            j = j + 1

t1 = time.time()

print(f'Method 1: {t1 - t0}')
res1 = df1.head()

df2 = pd.DataFrame({'price': data})
t2 = time.time()

count = len(df2)
df2['updown'] = np.nan
up = df2.price + 2
down = df2.price - 2

# increase shift range until updown is set for all columns
# or there is insufficient data to change remaining rows
i = 1
while (i < count) and (not (isna := df2.updown.isna()) is None and ((i == 1) or (isna[:-(i - 1)].any()))):
    shift = df2.price.shift(-i)
    df2.loc[isna & (shift >= up), 'updown'] =  'Up'
    df2.loc[isna & (shift <= down), 'updown'] = 'Down'
    i += 1

t3 = time.time()
print(f'Method 2: {t3 - t2}')

s1 = df1.updown
s2 = df2.updown

match = (s1.isnull() == s2.isnull()).all() and (s1[s1.notnull()] == s2[s2.notnull()]).all()

print(f'Series match: {match}')

速度提高的主要原因是，我们在数据数组上进行操作，而不是在Python中遍历各行，这在C代码中都会发生.虽然Python调用Pandas或NumPy(这是C库)非常快，但有一些开销，如果您这样做的时间很长，它很快就会成为限制因素.

性能的提高取决于输入数据，但随数据帧中的行数而扩展:行越多，迭代速度就越慢:

   iterations     method1   method2     increase
0         100    0.056002  0.018267     3.065689
1        1000    0.209895  0.005000    41.982070
2       10000    2.625701  0.009001   291.727054
3      100000  108.080149  0.042001  2573.260448