Python 3.x 如何通过替换pandas中的for循环来减少程序执行时间

我有12000个csv文件，每个文件有6000行。我在代码中使用for循环，我认为正因为如此，我的代码执行时间增加了。如果有人知道如何将这段代码更改到pandas包中以减少执行时间

for i in range(len(df)):
        if ((df['EOG_Start_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_start_farm'].values[i]-df['EOG_Min_Farm'].values[i])) &((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
          #print('EOG')
          df['EOG_flag'].values[i]=1

        if ((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
            #print('gust')
            df['Gust_flag'].values[i]=1

范围内的i（len（df））：
如果（（df['EOG\u Start\u model'].values[i]-df['EOG\u Min\u model'].values[i]）您可以单独使用矢量化解决方案-craete布尔掩码，通过
和
链接在一起，并在以下位置设置值：

x=df['EOG\u start\u farm']。值df['EOG\u Min\u farm']。值
m1=（df['EOG\U Start\U model']值df['EOG\U Min\U model']值）小于x
m2=（df['EOG_最大_模型]]值df['EOG_最小_模型]]值）小于x
m3=df['Avg']。数值>2
m23=m2和m3
df['EOG_标志']=np.式中（m1&m2&m3，1，df['EOG_标志'].值）
df['Gust_flag']=np.式中（m2和m3，1，df['Gust_flag']值）

性能：

np.random.seed(2019)

N = 6000
c = ['EOG_Start_model','EOG_Min_model','EOG_start_farm','EOG_Min_Farm','EOG_Max_model',
     'EOG_Max_Farm','Avg','EOG_flag','Gust_flag']
df = pd.DataFrame(np.random.rand(N, 9), columns=c)
print (df)

In [91]: %%timeit
    ...: x = df['EOG_start_farm'].values-df['EOG_Min_Farm'].values
    ...: m1 = (df['EOG_Start_model'].values-df['EOG_Min_model'].values) < x
    ...: m2 = (df['EOG_Max_model'].values-df['EOG_Min_model'].values) < x
    ...: m3 = df['Avg'].values > 2
    ...: m23 = m2 & m3
    ...: 
    ...: df['EOG_flag'] = np.where(m1 & m2 & m3, 1, df['EOG_flag'].values)
    ...: df['Gust_flag'] = np.where(m2 & m3, 1, df['Gust_flag'].values)
    ...: 
597 µs ± 6.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [93]: %%timeit
    ...: for i in range(len(df)):
    ...:     if ((df['EOG_Start_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_start_farm'].values[i]-df['EOG_Min_Farm'].values[i])) &((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
    ...:           #print('EOG')
    ...:           df['EOG_flag'].values[i]=1
    ...: 
    ...:     if ((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
    ...:             #print('gust')
    ...:             df['Gust_flag'].values[i]=1
231 ms ± 1.16 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

np.random.seed（2019年）
N=6000
c=[‘EOG_开始_模型’、‘EOG_最小_模型’、‘EOG_开始_农场’、‘EOG_最小_农场’、‘EOG_最大_模型’，
‘EOG_Max_Farm’、‘Avg’、‘EOG_flag’、‘Gust_flag’]
df=pd.DataFrame（np.random.rand（N，9），columns=c）
打印（df）
在[91]中：%%timeit
…：x=df['EOG\u start\u farm']。值df['EOG\u Min\u farm']。值
…：m1=（df['EOG_Start_model']值df['EOG_Min_model']值）2
…：m23=m2和m3
...: 
…：df['EOG_flag']=np.式中（m1&m2&m3，1，df['EOG_flag']值）
…：df['Gust_flag']=np.式中（m2和m3，1，df['Gust_flag']值）
...: 
每个回路597µs±6.3µs（7次运行的平均值±标准偏差，每个1000个回路）
在[93]中：%%timeit
…：对于范围内的i（len（df））：
…：如果（（df['EOG\u Start\u model'].values[i]-df['EOG\u Min\u model'].values[i]）此代码与for循环具有相同的结果，则没有明显的执行时间，执行几乎相同time@Nickel-好的，添加一些测试。@Nickel-它比原始溶液快387倍。
np.random.seed(2019)

N = 6000
c = ['EOG_Start_model','EOG_Min_model','EOG_start_farm','EOG_Min_Farm','EOG_Max_model',
     'EOG_Max_Farm','Avg','EOG_flag','Gust_flag']
df = pd.DataFrame(np.random.rand(N, 9), columns=c)
print (df)

In [91]: %%timeit
    ...: x = df['EOG_start_farm'].values-df['EOG_Min_Farm'].values
    ...: m1 = (df['EOG_Start_model'].values-df['EOG_Min_model'].values) < x
    ...: m2 = (df['EOG_Max_model'].values-df['EOG_Min_model'].values) < x
    ...: m3 = df['Avg'].values > 2
    ...: m23 = m2 & m3
    ...: 
    ...: df['EOG_flag'] = np.where(m1 & m2 & m3, 1, df['EOG_flag'].values)
    ...: df['Gust_flag'] = np.where(m2 & m3, 1, df['Gust_flag'].values)
    ...: 
597 µs ± 6.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [93]: %%timeit
    ...: for i in range(len(df)):
    ...:     if ((df['EOG_Start_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_start_farm'].values[i]-df['EOG_Min_Farm'].values[i])) &((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
    ...:           #print('EOG')
    ...:           df['EOG_flag'].values[i]=1
    ...: 
    ...:     if ((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
    ...:             #print('gust')
    ...:             df['Gust_flag'].values[i]=1
231 ms ± 1.16 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)