Python 使用DataFrame创建三个新列

我在下面有一个数据框，并尝试创建三个新的列“大”、“小”和“计数”。条件是计算有多少值大于/小于平均值并求和

df = 
            APPL       Std_1       Std_2       Std_3          Mean
       0   ACCMGR      106.8754    130.1600    107.1861    114.750510
       1   ACCOUNTS    121.7034    113.4927    114.5482    116.581458
       2   AUTH        116.8585    112.4487    115.2700    114.859050

def make_count(comp_cols, mean_col):
    count_d = {'greater': 0, 'less': 0}
    for col in comp_cols:
        if col > mean_col:
            count_d['greater'] += 1
        elif col < mean_col:
            count_d['less'] += 1
    return count_d['greater'], count_d['less'], (count_d['greater'] + count_d['less'])


def apply_make_count(df):
    a,b,c,*d= df.apply(lambda row: make_count([row['Std_1'], row['Std_2'], row['Std_3']], row['Mean of Std']), axis=1)
    df['greater'],df['less'],df['count']=a,b,c

apply_make_count(df)

我想成为什么样的人

 df = 
    APPL       Std_1       Std_2       Std_3      Mean  greater less    count
0   ACCMGR      106.8754    130.1600    107.1861    114.750510        1    2        3
1   ACCOUNTS    121.7034    113.4927    114.5482    116.581458        1    2        3
2   AUTH        116.8585    112.4487    115.2700    114.859050        2    1        3

---

看来你只需要

sub_df = df[['Std_1', 'Std_2', 'Std_3']]

df['greater'] = sub_df.gt(df.Mean.values).sum(1) # same as (sub_df > df.Mean.values).sum(1)
df['less']    = sub_df.lt(df.Mean.values).sum(1)
df['count']   = sub_df.count(1)


    APPL        Std_1       Std_2       Std_3       Mean        greater less   count
0   ACCMGR      106.8754    130.1600    107.1861    114.750510  1       2      3
1   ACCOUNTS    121.7034    113.4927    114.5482    116.581458  1       2      3
2   AUTH        116.8585    112.4487    115.2700    114.859050  2       1      3

---

看来你只需要

sub_df = df[['Std_1', 'Std_2', 'Std_3']]

df['greater'] = sub_df.gt(df.Mean.values).sum(1) # same as (sub_df > df.Mean.values).sum(1)
df['less']    = sub_df.lt(df.Mean.values).sum(1)
df['count']   = sub_df.count(1)


    APPL        Std_1       Std_2       Std_3       Mean        greater less   count
0   ACCMGR      106.8754    130.1600    107.1861    114.750510  1       2      3
1   ACCOUNTS    121.7034    113.4927    114.5482    116.581458  1       2      3
2   AUTH        116.8585    112.4487    115.2700    114.859050  2       1      3

试一试

df['greater']=（df.iloc[：，1:4].值>df['Mean']].值）.和（轴=1）
df['less']=（df.iloc[：，1:4）.值

试试看

df['greater']=（df.iloc[：，1:4].值>df['Mean']].值）.和（轴=1）
df['less']=（df.iloc[：，1:4）.值

您会收到错误，因为您在提供的原始解决方案中添加了

，*d

# the way you rewrote it
a,b,c,*d= df.apply(lambda row: make_count([row['Std_1'], row['Std_2'], row['Std_3']], row['Mean of Std']), axis=1)
df['greater'], df['less'], df['count'] = a, b, c

# the code you were provided
a, b, c = df.apply(lambda row: make_count([row['Std_1'], row['Std_2'], row['Std_3']], row['Mean']), axis=1)
df['greater'], df['less'], df['count'] = list(zip(a, b, c))

您在此提供的解决方案

此外：

提供给您的原始解决方案是最快的解决方案：

%timeit(apply_make_count(df))
1.93 ms ± 279 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

新的解决方案：

def test():
    df['greater'] = (df.iloc[:, 1:4].values > df[['Mean']].values).sum(axis=1)
    df['less'] = (df.iloc[:, 1:4].values < df[['Mean']].values).sum(axis=1)
    df['count'] = df.iloc[:, 1:4].count(1)

%timeit(test())
2.6 ms ± 35.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)


def test2():
    sub_df = df[['Std_1', 'Std_2', 'Std_3']]
    df['greater'] = sub_df.gt(df.Mean.values).sum(1) # same as (sub_df > df.Mean.values).sum(1)
    df['less']    = sub_df.lt(df.Mean.values).sum(1)
    df['count']   = sub_df.count(1)

%timeit(test2())
2.82 ms ± 263 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

def test（）：
df['greater']=（df.iloc[：，1:4]。值>df[['Mean']]。值）。总和（轴=1）
df['less']=（df.iloc[：，1:4）.值df.Mean.values）相同。总和（1）
df['less']=sub_df.lt（df.平均值）.sum（1）
df['count']=子函数df.count（1）
%timeit（test2（））
每个回路2.82 ms±263µs（7次运行的平均值±标准偏差，每个100个回路）

您会收到错误，因为您在提供的原始解决方案中添加了

，*d

# the way you rewrote it
a,b,c,*d= df.apply(lambda row: make_count([row['Std_1'], row['Std_2'], row['Std_3']], row['Mean of Std']), axis=1)
df['greater'], df['less'], df['count'] = a, b, c

# the code you were provided
a, b, c = df.apply(lambda row: make_count([row['Std_1'], row['Std_2'], row['Std_3']], row['Mean']), axis=1)
df['greater'], df['less'], df['count'] = list(zip(a, b, c))

您在此提供的解决方案

此外：

提供给您的原始解决方案是最快的解决方案：

%timeit(apply_make_count(df))
1.93 ms ± 279 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

新的解决方案：

def test():
    df['greater'] = (df.iloc[:, 1:4].values > df[['Mean']].values).sum(axis=1)
    df['less'] = (df.iloc[:, 1:4].values < df[['Mean']].values).sum(axis=1)
    df['count'] = df.iloc[:, 1:4].count(1)

%timeit(test())
2.6 ms ± 35.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)


def test2():
    sub_df = df[['Std_1', 'Std_2', 'Std_3']]
    df['greater'] = sub_df.gt(df.Mean.values).sum(1) # same as (sub_df > df.Mean.values).sum(1)
    df['less']    = sub_df.lt(df.Mean.values).sum(1)
    df['count']   = sub_df.count(1)

%timeit(test2())
2.82 ms ± 263 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

def test（）：
df['greater']=（df.iloc[：，1:4]。值>df[['Mean']]。值）。总和（轴=1）
df['less']=（df.iloc[：，1:4）.值df.Mean.values）相同。总和（1）
df['less']=sub_df.lt（df.平均值）.sum（1）
df['count']=子函数df.count（1）
%timeit（test2（））
每个回路2.82 ms±263µs（7次运行的平均值±标准偏差，每个100个回路）