Pandas - 9.2 向量化函数

import pandas as pd

df = pd.DataFrame({'a': [10, 20, 30],
                   'b': [20, 30, 40]})
print(df)
'''
    a   b
0  10  20
1  20  30
2  30  40
'''

# avg_2函数内部的计算本质上是向量化的
def avg_2(x, y):
    return (x + y)/2

print(avg_2(df['a'], df['b']))

'''
0    15.0
1    25.0
2    35.0
dtype: float64
'''

import numpy as np

# 非向量化计算
def avg_2_mod(x, y):
    if x == 20:
        return np.NaN
    else:
        return (x + y)/2

print(avg_2_mod(df['a'], df['b']))
# 报错
# ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

# 输入单个值正常工作
print(avg_2_mod(10, 20))
# 15.0

np.vectorize 函数

对于非向量化的函数，使用np.vectorize创建新函数实现向量化(对于没有某个函数的源代码时)

avg_2_mod_vec = np.vectorize(avg_2_mod)
print(avg_2_mod_vec(df['a'], df['b']))
# [15. nan 35.]

python装饰器把函数向量化，无需创建新函数（对于函数是自定义的）

@np.vectorize
def avg_2_mod(x, y):
    if x == 20:
        return np.NaN
    else:
        return (x + y)/2

print(avg_2_mod(df['a'], df['b']))
# [15. nan 35.]

lambda 函数

当函数相当简单时，可以写在apply方法中

# 编写一个模式，从数据行中提取所有字母，并把它们赋给新的列name。
import regex

p = regex.compile('\w+\s+\w+')

def get_name(s):
    return p.match(s).group()

docs = pd.read_csv('data/doctors.csv', header=None)
docs['name_func'] = docs[0].apply(get_name)
print(docs)

'''
                               0              name_func
0     William Hartnell (1963-66)       William Hartnell
1    Patrick Troughton (1966-69)      Patrick Troughton
2          Jon Pertwee (1970 74)            Jon Pertwee
3            Tom Baker (1974-81)              Tom Baker
4        Peter Davison (1982-84)          Peter Davison
5          Colin Baker (1984-86)            Colin Baker
6      Sylvester McCoy (1987-89)        Sylvester McCoy
7             Paul McGann (1996)            Paul McGann
8   Christopher Eccleston (2005)  Christopher Eccleston
9        David Tennant (2005-10)          David Tennant
10          Matt Smith (2010-13)             Matt Smith
11     Peter Capaldi (2014-2017)          Peter Capaldi
12        Jodie Whittaker (2017)        Jodie Whittaker
'''

# 函数简单，直接写在apply中，lambda会将整列或整行作为第一个参数
docs['name_lamb'] = docs[0].apply(lambda x: p.match(x).group())
print(docs)

'''
                               0              name_func              name_lamb
0     William Hartnell (1963-66)       William Hartnell       William Hartnell
1    Patrick Troughton (1966-69)      Patrick Troughton      Patrick Troughton
2          Jon Pertwee (1970 74)            Jon Pertwee            Jon Pertwee
3            Tom Baker (1974-81)              Tom Baker              Tom Baker
4        Peter Davison (1982-84)          Peter Davison          Peter Davison
5          Colin Baker (1984-86)            Colin Baker            Colin Baker
6      Sylvester McCoy (1987-89)        Sylvester McCoy        Sylvester McCoy
7             Paul McGann (1996)            Paul McGann            Paul McGann
8   Christopher Eccleston (2005)  Christopher Eccleston  Christopher Eccleston
9        David Tennant (2005-10)          David Tennant          David Tennant
10          Matt Smith (2010-13)             Matt Smith             Matt Smith
11     Peter Capaldi (2014-2017)          Peter Capaldi          Peter Capaldi
12        Jodie Whittaker (2017)        Jodie Whittaker        Jodie Whittaker
'''