如何修改Python函数返回DataFrame中符合距离要求的词对位置列表？

你完全可以直接在函数中返回符合条件的i和j组成的元组列表来实现需求，只需要对原函数做简单修改：

修改后的完整代码 #

import pandas as pd

def identify_instances(string, word_1='can', word_2='help', allowed_distance=3):
    string_list = string.split(' ')
    indices_word_1 = [i for i, x in enumerate(string_list) if x == word_1]
    indices_word_2 = [i for i, x in enumerate(string_list) if x == word_2]
    # 将计数变量替换为列表，用于存储符合条件的位置对
    result = []
    for i in indices_word_1:
        for j in indices_word_2:
            _distance = abs(i - j)
            if _distance <= allowed_distance:
                # 符合条件时将位置对(i,j)添加到列表
                result.append((i, j))
    return result

# 测试示例数据
data = [['ABC123', 'can you help me with this'], 
        ['ABC456', 'I can not help with this, but they can help'], 
        ['ABC789', 'No one can help can they']]
pandasDF = pd.DataFrame(data, columns=['Record ID', 'String'])

# 新增列存储位置对列表
pandasDF['can-NEAR-help Positions'] = pandasDF['String'].apply(identify_instances)

# 输出目标结果
print(pandasDF[['Record ID', 'can-NEAR-help Positions']])

运行结果 #

Record ID can-NEAR-help Positions
0    ABC123                  [(0, 2)]
1    ABC456          [(1, 3), (8, 9)]
2    ABC789          [(2, 3), (3, 4)]

核心改动说明 #

• 把原函数中用于计数的result = 0改为空列表result = []
• 将计数操作result += 1替换为result.append((i, j))，直接存储符合距离要求的词对位置
• 最终返回存储所有有效位置对的列表

内容的提问来源于stack exchange，提问作者tshobe