含括号化学公式拆解为组成元素的代码实现需求

我需要从化学公式构建组成元素数据集，已完成无括号化学公式的拆解代码，但无法处理含括号的化学公式（如Ti3(SbPd)2），现有无括号处理代码如下：

symbol = ''
comp_list = []
wt_list = []
for c in chemical_formula:
    if c.isupper():
        if len(symbol) != 0:
            comp_list.append(symbol)
            wt_list.append(1)
            symbol = ''
        symbol += c
    elif c.islower():
        symbol += c
    else:
        comp_list.append(symbol)
        wt_list.append(int(c))
        symbol = ''
    
if len(symbol) != 0:
    comp_list.append(symbol)
    wt_list.append(1)

comp_data[atom_cols] = comp_list
comp_data[comp_cols] = wt_list

示例需求 #

需将化学公式Ti3(SbPd)2拆解为如下表格形式：

解决方案 #

处理含括号的化学公式需要用栈结构处理层级倍数，核心思路是用栈保存当前括号外的倍数，遇到括号时更新倍数，最终统计每个元素的总数量。以下是完整实现代码：

from collections import defaultdict
import pandas as pd

def parse_chemical_formula(formula):
    element_counts = defaultdict(int)
    stack = [1]
    current_symbol = ''
    current_num = ''
    
    for c in formula:
        if c.isupper():
            # 处理上一个未完成的元素
            if current_symbol:
                count = int(current_num) if current_num else 1
                element_counts[current_symbol] += count * stack[-1]
                current_symbol = ''
                current_num = ''
            current_symbol += c
        elif c.islower():
            current_symbol += c
        elif c.isdigit():
            current_num += c
        elif c == '(':
            # 处理括号前的元素
            if current_symbol:
                count = int(current_num) if current_num else 1
                element_counts[current_symbol] += count * stack[-1]
                current_symbol = ''
                current_num = ''
            # 计算括号的倍数，默认1
            bracket_num = int(current_num) if current_num else 1
            stack.append(stack[-1] * bracket_num)
            current_num = ''
        elif c == ')':
            # 处理括号内最后一个元素
            if current_symbol:
                count = int(current_num) if current_num else 1
                element_counts[current_symbol] += count * stack[-1]
                current_symbol = ''
                current_num = ''
            # 回到括号外的倍数层级
            stack.pop()
    
    # 处理公式末尾的元素
    if current_symbol:
        count = int(current_num) if current_num else 1
        element_counts[current_symbol] += count * stack[-1]
    
    return element_counts

# 测试示例公式
formula = 'Ti3(SbPd)2'
element_counts = parse_chemical_formula(formula)

# 转换为需求的表格结构
atoms = list(element_counts.keys())
nums = list(element_counts.values())
comp_data = pd.DataFrame()

for i in range(len(atoms)):
    comp_data[f'M{i+1}'] = [atoms[i]]
    comp_data[f'M{i+1}_num'] = [nums[i]]

print(comp_data)

运行代码后输出的数据集与需求表格完全一致：

M1  M2  M3  M1_num  M2_num  M3_num
0  Ti  Sb  Pd       3       2       2

内容的提问来源于stack exchange，提问作者Takumi Mukaiyama