基于Pandas DataFrame的字符块统计与条件计算问题

原始DataFrame定义 #

import pandas as pd
import re

please = pd.DataFrame({'name':['kitty', 'bake','kitty','bake'], 
                      'level' : ['12', '12', '15', '15'],
                      'col1':['......kkkkkkk......kkkkkkkkkkk.....kkkkkkkkkkkk', '......kkkkkkkkk.........kkkkkkk.......kkkkkkkk...', 'kkkk....kkkkkk....kkkkkkkkkkk......','kkkkkkk...kkkkkkkk.k.kkkkkkkkkk........kk']})   

原始DataFrame结构 #

name level                                               col1
0  kitty    12      ......kkkkkkk......kkkkkkkkkkk.....kkkkkkkkkkkk
1   bake    12  ......kkkkkkkkk.........kkkkkkk.......kkkkkkkk...
2  kitty    15                kkkk....kkkkkk....kkkkkkkkkkk......
3   bake    15               kkkkkkk...kkkkkkkk.k.kkkkkkkkkk........kk

问题1：统计每行col1中连续'k'字符块的长度列表 #

用正则匹配所有连续的'k'片段，提取每个片段的长度并生成新列：

# 提取连续k块的长度列表
def get_k_block_lengths(s):
    k_blocks = re.findall(r'k+', s)
    return [len(block) for block in k_blocks]

please['k_block_lengths'] = please['col1'].apply(get_k_block_lengths)

查看bake_15对应的结果：

print(please.loc[3, 'k_block_lengths'])

输出：[7,8,1,10,2]

问题2：过滤列表中小于5的数值 #

对生成的k块长度列表进行过滤，保留长度≥5的元素：

# 过滤函数
def filter_blocks(lst):
    return [x for x in lst if x >= 5]

please['k_block_lengths_filtered'] = please['k_block_lengths'].apply(filter_blocks)

查看bake_15过滤后的结果：

print(please.loc[3, 'k_block_lengths_filtered'])

输出：[7,8,10]

问题3：按条件提取指定位置的字符块长度 #

获取kitty_level_12的第2个'k'字符块长度 #

定位目标行后，直接提取对应k块的第2个元素（列表索引从0开始）：

kitty_12_row = please[(please['name'] == 'kitty') & (please['level'] == '12')].iloc[0]
kitty_12_2nd_k = kitty_12_row['k_block_lengths'][1]
print(f"kitty_level_12_2nd 'k' block = {kitty_12_2nd_k}")

输出：kitty_level_12_2nd 'k' block = 11

获取bake_level_12的第2个'.'字符块长度 #

先定义提取连续'.'块长度的函数，再定位目标行提取对应元素：

# 提取连续.块长度的函数
def get_dot_block_lengths(s):
    dot_blocks = re.findall(r'\.+', s)
    return [len(block) for block in dot_blocks]

bake_12_row = please[(please['name'] == 'bake') & (please['level'] == '12')].iloc[0]
bake_12_2nd_dot = get_dot_block_lengths(bake_12_row['col1'])[1]
print(f"bake_level_12_2nd '.' block = {bake_12_2nd_dot}")

输出：bake_level_12_2nd '.' block = 9

内容的提问来源于stack exchange，提问作者888Seeme