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358 lines
12 KiB
Python
358 lines
12 KiB
Python
import networkx as nx
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import re
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from typing import List, Dict, Any, Tuple, Union
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from .pump_protocol import generate_pump_protocol_with_rinsing
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def parse_volume_with_units(volume_input: Union[str, float, int], default_unit: str = "mL") -> float:
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"""
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解析带单位的体积输入
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Args:
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volume_input: 体积输入(如 "100 mL", "2.5 L", "500", "?", 100.0)
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default_unit: 默认单位(默认为毫升)
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Returns:
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float: 体积(毫升)
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"""
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if not volume_input:
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return 0.0
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# 处理数值输入
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if isinstance(volume_input, (int, float)):
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result = float(volume_input)
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print(f"RECRYSTALLIZE: 数值体积输入: {volume_input} → {result}mL(默认单位)")
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return result
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# 处理字符串输入
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volume_str = str(volume_input).lower().strip()
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print(f"RECRYSTALLIZE: 解析体积字符串: '{volume_str}'")
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# 处理特殊值
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if volume_str in ['?', 'unknown', 'tbd', 'to be determined']:
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default_volume = 50.0 # 50mL默认值
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print(f"RECRYSTALLIZE: 检测到未知体积,使用默认值: {default_volume}mL")
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return default_volume
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# 如果是纯数字,使用默认单位
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try:
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value = float(volume_str)
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if default_unit.lower() in ["ml", "milliliter"]:
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result = value
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elif default_unit.lower() in ["l", "liter"]:
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result = value * 1000.0
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elif default_unit.lower() in ["μl", "ul", "microliter"]:
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result = value / 1000.0
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else:
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result = value # 默认mL
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print(f"RECRYSTALLIZE: 纯数字输入: {volume_str} → {result}mL(单位: {default_unit})")
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return result
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except ValueError:
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pass
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# 移除空格并提取数字和单位
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volume_clean = re.sub(r'\s+', '', volume_str)
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# 匹配数字和单位的正则表达式
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match = re.match(r'([0-9]*\.?[0-9]+)\s*(ml|l|μl|ul|microliter|milliliter|liter)?', volume_clean)
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if not match:
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print(f"RECRYSTALLIZE: ⚠️ 无法解析体积: '{volume_str}',使用默认值: 50mL")
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return 50.0
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value = float(match.group(1))
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unit = match.group(2) or default_unit.lower()
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# 转换为毫升
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if unit in ['l', 'liter']:
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volume = value * 1000.0 # L -> mL
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elif unit in ['μl', 'ul', 'microliter']:
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volume = value / 1000.0 # μL -> mL
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else: # ml, milliliter 或默认
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volume = value # 已经是mL
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print(f"RECRYSTALLIZE: 体积解析: '{volume_str}' → {value} {unit} → {volume}mL")
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return volume
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def parse_ratio(ratio_str: str) -> Tuple[float, float]:
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"""
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解析比例字符串,支持多种格式
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Args:
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ratio_str: 比例字符串(如 "1:1", "3:7", "50:50")
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Returns:
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Tuple[float, float]: 比例元组 (ratio1, ratio2)
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"""
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try:
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# 处理 "1:1", "3:7", "50:50" 等格式
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if ":" in ratio_str:
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parts = ratio_str.split(":")
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if len(parts) == 2:
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ratio1 = float(parts[0])
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ratio2 = float(parts[1])
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return ratio1, ratio2
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# 处理 "1-1", "3-7" 等格式
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if "-" in ratio_str:
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parts = ratio_str.split("-")
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if len(parts) == 2:
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ratio1 = float(parts[0])
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ratio2 = float(parts[1])
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return ratio1, ratio2
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# 处理 "1,1", "3,7" 等格式
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if "," in ratio_str:
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parts = ratio_str.split(",")
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if len(parts) == 2:
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ratio1 = float(parts[0])
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ratio2 = float(parts[1])
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return ratio1, ratio2
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# 默认 1:1
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print(f"RECRYSTALLIZE: 无法解析比例 '{ratio_str}',使用默认比例 1:1")
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return 1.0, 1.0
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except ValueError:
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print(f"RECRYSTALLIZE: 比例解析错误 '{ratio_str}',使用默认比例 1:1")
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return 1.0, 1.0
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def find_solvent_vessel(G: nx.DiGraph, solvent: str) -> str:
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"""
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查找溶剂容器
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Args:
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G: 网络图
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solvent: 溶剂名称
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Returns:
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str: 溶剂容器ID
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"""
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print(f"RECRYSTALLIZE: 正在查找溶剂 '{solvent}' 的容器...")
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# 构建可能的容器名称
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possible_names = [
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f"flask_{solvent}",
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f"bottle_{solvent}",
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f"reagent_{solvent}",
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f"reagent_bottle_{solvent}",
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f"{solvent}_flask",
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f"{solvent}_bottle",
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f"{solvent}",
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f"vessel_{solvent}",
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]
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# 第一步:通过容器名称匹配
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for vessel_name in possible_names:
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if vessel_name in G.nodes():
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print(f"RECRYSTALLIZE: 通过名称匹配找到容器: {vessel_name}")
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return vessel_name
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# 第二步:通过模糊匹配
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for node_id in G.nodes():
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if G.nodes[node_id].get('type') == 'container':
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node_name = G.nodes[node_id].get('name', '').lower()
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if solvent.lower() in node_id.lower() or solvent.lower() in node_name:
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print(f"RECRYSTALLIZE: 通过模糊匹配找到容器: {node_id}")
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return node_id
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# 第三步:通过液体类型匹配
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for node_id in G.nodes():
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if G.nodes[node_id].get('type') == 'container':
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vessel_data = G.nodes[node_id].get('data', {})
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liquids = vessel_data.get('liquid', [])
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for liquid in liquids:
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if isinstance(liquid, dict):
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liquid_type = (liquid.get('liquid_type') or liquid.get('name', '')).lower()
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reagent_name = vessel_data.get('reagent_name', '').lower()
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if solvent.lower() in liquid_type or solvent.lower() in reagent_name:
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print(f"RECRYSTALLIZE: 通过液体类型匹配找到容器: {node_id}")
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return node_id
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raise ValueError(f"找不到溶剂 '{solvent}' 对应的容器")
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def generate_recrystallize_protocol(
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G: nx.DiGraph,
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ratio: str,
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solvent1: str,
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solvent2: str,
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vessel: str,
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volume: Union[str, float], # 🔧 修改:支持字符串和数值
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**kwargs
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) -> List[Dict[str, Any]]:
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"""
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生成重结晶协议序列 - 支持单位
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Args:
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G: 有向图,节点为容器和设备
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ratio: 溶剂比例(如 "1:1", "3:7")
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solvent1: 第一种溶剂名称
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solvent2: 第二种溶剂名称
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vessel: 目标容器
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volume: 总体积(支持 "100 mL", "50", "2.5 L" 等)
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**kwargs: 其他可选参数
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Returns:
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List[Dict[str, Any]]: 动作序列
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"""
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action_sequence = []
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print(f"RECRYSTALLIZE: 开始生成重结晶协议(支持单位)")
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print(f" - 比例: {ratio}")
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print(f" - 溶剂1: {solvent1}")
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print(f" - 溶剂2: {solvent2}")
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print(f" - 容器: {vessel}")
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print(f" - 总体积: {volume} (类型: {type(volume)})")
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# 1. 验证目标容器存在
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if vessel not in G.nodes():
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raise ValueError(f"目标容器 '{vessel}' 不存在于系统中")
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# 2. 🔧 新增:解析体积(支持单位)
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final_volume = parse_volume_with_units(volume, "mL")
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print(f"RECRYSTALLIZE: 解析体积: {volume} → {final_volume}mL")
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# 3. 解析比例
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ratio1, ratio2 = parse_ratio(ratio)
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total_ratio = ratio1 + ratio2
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# 4. 计算各溶剂体积
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volume1 = final_volume * (ratio1 / total_ratio)
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volume2 = final_volume * (ratio2 / total_ratio)
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print(f"RECRYSTALLIZE: 解析比例: {ratio1}:{ratio2}")
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print(f"RECRYSTALLIZE: {solvent1} 体积: {volume1:.2f} mL")
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print(f"RECRYSTALLIZE: {solvent2} 体积: {volume2:.2f} mL")
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# 5. 查找溶剂容器
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try:
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solvent1_vessel = find_solvent_vessel(G, solvent1)
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print(f"RECRYSTALLIZE: 找到溶剂1容器: {solvent1_vessel}")
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except ValueError as e:
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raise ValueError(f"无法找到溶剂1 '{solvent1}': {str(e)}")
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try:
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solvent2_vessel = find_solvent_vessel(G, solvent2)
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print(f"RECRYSTALLIZE: 找到溶剂2容器: {solvent2_vessel}")
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except ValueError as e:
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raise ValueError(f"无法找到溶剂2 '{solvent2}': {str(e)}")
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# 6. 验证路径存在
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try:
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path1 = nx.shortest_path(G, source=solvent1_vessel, target=vessel)
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print(f"RECRYSTALLIZE: 溶剂1路径: {' → '.join(path1)}")
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except nx.NetworkXNoPath:
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raise ValueError(f"从溶剂1容器 '{solvent1_vessel}' 到目标容器 '{vessel}' 没有可用路径")
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try:
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path2 = nx.shortest_path(G, source=solvent2_vessel, target=vessel)
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print(f"RECRYSTALLIZE: 溶剂2路径: {' → '.join(path2)}")
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except nx.NetworkXNoPath:
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raise ValueError(f"从溶剂2容器 '{solvent2_vessel}' 到目标容器 '{vessel}' 没有可用路径")
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# 7. 添加第一种溶剂
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print(f"RECRYSTALLIZE: 开始添加溶剂1 {volume1:.2f} mL")
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try:
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pump_actions1 = generate_pump_protocol_with_rinsing(
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G=G,
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from_vessel=solvent1_vessel,
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to_vessel=vessel,
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volume=volume1, # 使用解析后的体积
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amount="",
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time=0.0,
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viscous=False,
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rinsing_solvent="", # 重结晶不需要清洗
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rinsing_volume=0.0,
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rinsing_repeats=0,
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solid=False,
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flowrate=2.0, # 正常流速
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transfer_flowrate=0.5
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)
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action_sequence.extend(pump_actions1)
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except Exception as e:
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raise ValueError(f"生成溶剂1泵协议时出错: {str(e)}")
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# 8. 等待溶剂1稳定
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action_sequence.append({
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"action_name": "wait",
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"action_kwargs": {
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"time": 10.0,
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"description": f"等待溶剂1 {solvent1} 稳定"
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}
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})
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# 9. 添加第二种溶剂
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print(f"RECRYSTALLIZE: 开始添加溶剂2 {volume2:.2f} mL")
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try:
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pump_actions2 = generate_pump_protocol_with_rinsing(
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G=G,
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from_vessel=solvent2_vessel,
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to_vessel=vessel,
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volume=volume2, # 使用解析后的体积
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amount="",
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time=0.0,
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viscous=False,
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rinsing_solvent="", # 重结晶不需要清洗
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rinsing_volume=0.0,
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rinsing_repeats=0,
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solid=False,
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flowrate=2.0, # 正常流速
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transfer_flowrate=0.5
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)
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action_sequence.extend(pump_actions2)
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except Exception as e:
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raise ValueError(f"生成溶剂2泵协议时出错: {str(e)}")
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# 10. 等待溶剂2稳定
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action_sequence.append({
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"action_name": "wait",
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"action_kwargs": {
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"time": 10.0,
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"description": f"等待溶剂2 {solvent2} 稳定"
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}
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})
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# 11. 等待重结晶完成
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action_sequence.append({
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"action_name": "wait",
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"action_kwargs": {
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"time": 600.0, # 等待10分钟进行重结晶
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"description": f"等待重结晶完成({solvent1}:{solvent2} = {ratio},总体积 {final_volume}mL)"
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}
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})
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print(f"RECRYSTALLIZE: 协议生成完成,共 {len(action_sequence)} 个动作")
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print(f"RECRYSTALLIZE: 预计总时间: {620/60:.1f} 分钟")
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print(f"RECRYSTALLIZE: 总体积: {final_volume}mL")
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return action_sequence
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# 测试函数
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def test_recrystallize_protocol():
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"""测试重结晶协议"""
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print("=== RECRYSTALLIZE PROTOCOL 测试 ===")
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# 测试比例解析
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test_ratios = ["1:1", "3:7", "50:50", "1-1", "2,8", "invalid"]
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for ratio in test_ratios:
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r1, r2 = parse_ratio(ratio)
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print(f"比例 '{ratio}' -> {r1}:{r2}")
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print("测试完成")
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if __name__ == "__main__":
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test_recrystallize_protocol() |