运行列表数据框列中每个元素的函数

Question

这个对我来说有点棘手。

数据框：

parent           children
0   MAX          [MAX, amx, akd]
1   Sam          ['Sam','sammy','samsam']
2   Larry        ['lar','lair','larrylamo']

我有一个函数，如果我只传入一个字符串，它会比较这两个字符串，并打印出一个描述字符（距离）有多近的数字。类似于 levenshtein 方程。

如何在数据帧上运行这个函数？我需要将第一列 ('parent) 中的每条记录与第二列 ('children')?

中的相应列表进行比较

目前，我可以运行得到这个结果：

>>> reference = 'larry'
>>> value_list = ['lar','lair','larrylamo']
>>> get_top_matches(reference,value_list)
>>> [('lar',0.91),('larrylamo',0.91),('lair',0.83)]

我正在尝试为匹配的每一行创建由元组组成的第三列，如下所示：

parent           children                     func_results
0   MAX          [MAX, amx, akd]              [('MAX',1.0),('amx',0.89),('akd',0.56)]
1   Sam          ['Sam','sammy','samsam']     [('Sam',1.0),('sammy',0.91), ('samsam',0.88)]
2   Larry        ['lar','lair','larrylamo']   [('lar',0.91),('larrylamo',0.91), ('lair',0.83)]

我认为如果我能弄清楚如何在针对 df 的 for 循环中应用它，该函数应该能够按原样工作。

函数如下：

import math
import re

def sort_token_alphabetically(word):
    token = re.split('[,. ]', word)
    sorted_token = sorted(token)
    return ' '.join(sorted_token)

def get_jaro_distance(first, second, winkler=True, winkler_ajustment=True,
                      scaling=0.1, sort_tokens=True):
    if sort_tokens:
        first = sort_token_alphabetically(first)
        second = sort_token_alphabetically(second)

    if not first or not second:
        raise JaroDistanceException(
            "Cannot calculate distance from NoneType ({0}, {1})".format(
                first.__class__.__name__,
                second.__class__.__name__))

    jaro = _score(first, second)
    cl = min(len(_get_prefix(first, second)), 4)

    if all([winkler, winkler_ajustment]):  # 0.1 as scaling factor
        return round((jaro + (scaling * cl * (1.0 - jaro))) * 100.0) / 100.0

    return jaro

def _score(first, second):
    shorter, longer = first.lower(), second.lower()

    if len(first) > len(second):
        longer, shorter = shorter, longer

    m1 = _get_matching_characters(shorter, longer)
    m2 = _get_matching_characters(longer, shorter)

    if len(m1) == 0 or len(m2) == 0:
        return 0.0

    return (float(len(m1)) / len(shorter) +
            float(len(m2)) / len(longer) +
            float(len(m1) - _transpositions(m1, m2)) / len(m1)) / 3.0

def _get_diff_index(first, second):
    if first == second:
        pass

    if not first or not second:
        return 0

    max_len = min(len(first), len(second))
    for i in range(0, max_len):
        if not first[i] == second[i]:
            return i

    return max_len

def _get_prefix(first, second):
    if not first or not second:
        return ""

    index = _get_diff_index(first, second)
    if index == -1:
        return first

    elif index == 0:
        return ""

    else:
        return first[0:index]

def _get_matching_characters(first, second):
    common = []
    limit = math.floor(min(len(first), len(second)) / 2)

    for i, l in enumerate(first):
        left, right = int(max(0, i - limit)), int(
            min(i + limit + 1, len(second)))
        if l in second[left:right]:
            common.append(l)
            second = second[0:second.index(l)] + '*' + second[
                                                       second.index(l) + 1:]

    return ''.join(common)

def _transpositions(first, second):
    return math.floor(
        len([(f, s) for f, s in zip(first, second) if not f == s]) / 2.0)

def get_top_matches(reference, value_list, max_results=None):
    scores = []
    if not max_results:
        max_results = len(value_list)
    for val in value_list:
        score_sorted = get_jaro_distance(reference, val)
        score_unsorted = get_jaro_distance(reference, val, sort_tokens=False)
        scores.append((val, max(score_sorted, score_unsorted)))
    scores.sort(key=lambda x: x[1], reverse=True)

    return scores[:max_results]

class JaroDistanceException(Exception):
    def __init__(self, message):
        super(Exception, self).__init__(message)


reference = 'larry'
value_list = ['lar','lair','larrylamo']
get_top_matches(reference, value_list)

Answer 1

我假设您的真实数据集恰好有 2 列作为您的样本。在 axis=1

上使用 agg

df['func_results'] = df.agg(lambda x: get_top_matches(*x), axis=1)


Out[366]:
  parent  ...                                    func_results
0    MAX  ...          [(MAX, 1.0), (amx, 0.89), (akd, 0.56)]
1    Sam  ...     [(Sam, 1.0), (sammy, 0.87), (samsam, 0.83)]
2  Larry  ...  [(lar, 0.87), (larrylamo, 0.85), (lair, 0.78)]

[3 rows x 3 columns]

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