对Pandas系列应用多次迭代函数的最佳方法

Question

我需要对具有许多不同迭代的pandas系列应用带条件的文本替换。实现这一目标的最佳方式是什么

我的第一个想法是定义一个函数，在函数中迭代，然后应用它，但是这显然不起作用，因为每行只返回一个值（仅第一次迭代）：

def numberreplace(x):
    matches = {'FIRST':'1ST',
               'SECOND':'2ND',
               'THIRD':'3RD',
               'FOURTH':'4TH',
               'FIFTH':'5TH',
               'SIXTH':'6TH',
               'SEVENTH':'7TH',
               'EIGTH':'8TH',
               'NINTH':'9TH',
               'TENTH':'10TH'}
    for key in matches.keys():
        if (' '+key+'' in x) or (x.startswith(key)):
            x = x.replace(key, matches[key])
            return x
        else:
            return x
data['STREET REFORMAT'] = data['STREET REFORMAT'].apply(numberreplace)

我的另一个想法是在apply语句之外定义一个列表，迭代字典键列表，然后使用list元素将函数应用于行，但是，我不确定如何将多参数函数应用于一个系列，以及指定哪个参数是该系列的“row”参数

def numberreplace(row,k):
    matches = {'FIRST':'1ST',
                   'SECOND':'2ND',
                   'THIRD':'3RD',
                   'FOURTH':'4TH',
                   'FIFTH':'5TH',
                   'SIXTH':'6TH',
                   'SEVENTH':'7TH',
                   'EIGTH':'8TH',
                   'NINTH':'9TH',
                   'TENTH':'10TH'}
    if (' '+k+'' in row) or (row.startswith(k)):
        row = row.replace(k, matches[k])
        return row
    return row
    
    nummatches = ['FIRST','SECOND','THIRD','FOURTH','FIFTH','SIXTH','SEVENTH','EIGHTH','NINTH','TENTH]
    for match in nummatches:
        data['STREET REFORMAT'] = data['STREET REFORMAT'].apply(numberreplace(match))
        

在具有多行和多个替换字符串的数据帧上运行此apply函数最有效的方法是什么

Answer 1

重写函数的一种方法是使用正则表达式：

import re
def numberreplace(x):
    matches = {'FIRST':'1ST',
               'SECOND':'2ND',
               'THIRD':'3RD',
               'FOURTH':'4TH',
               'FIFTH':'5TH',
               'SIXTH':'6TH',
               'SEVENTH':'7TH',
               'EIGTH':'8TH',
               'NINTH':'9TH',
               'TENTH':'10TH'}
    for key in matches.keys():
        x = re.sub(re.compile(f"\s*{key}"), matches[key], x)
    return x
data['STREET REFORMAT'] = data['STREET REFORMAT'].apply(numberreplace)

这将用等价的matches替换任何出现的keys，如果找不到匹配项，则返回原始字符串。此解决方案比使用string.replace()方法的解决方案快约2倍，因此它可能对具有许多行和许多替换字符串的大型数据帧有用

Answer 2

代替循环，您可以创建两个条件并使用字典执行mask：

data = pd.DataFrame({"STREET REFORMAT":["FIRST", "THIRD", "IAMNINTH", "EIGTHISME"]})

cond1 = data["STREET REFORMAT"].str.contains("|".join(fr"\b{i}\b" for i in matches))
cond2 = data["STREET REFORMAT"].str.contains("|".join(fr"^{i}" for i in matches))

print (data["STREET REFORMAT"].mask(cond1|cond2, data["STREET REFORMAT"].replace(matches, regex=True)))

0         1ST
1         3RD
2    IAMNINTH
3     8THISME
Name: STREET REFORMAT, dtype: object

Answer 3

不需要在for循环中返回x，只需要在循环结束后返回一次。试试这个：

def numberreplace(x):
    matches = {'FIRST':'1ST',
               'SECOND':'2ND',
               'THIRD':'3RD',
               'FOURTH':'4TH',
               'FIFTH':'5TH',
               'SIXTH':'6TH',
               'SEVENTH':'7TH',
               'EIGTH':'8TH',
               'NINTH':'9TH',
               'TENTH':'10TH'}
    for key in matches.keys():
        if (' '+key+'' in x) or (x.startswith(key)):
            x = x.replace(key, matches[key])
    return x
data['STREET REFORMAT'] = data['STREET REFORMAT'].apply(numberreplace)