内核在使用纸浆求解器的 Jupyter 笔记本中不断死亡

Question

我在 Jupyter 笔记本中创建了一个 LP 求解器，这给了我一些问题。 具体来说，当我在下面的脚本中运行最后一行代码时，我收到错误消息说The kernel appears to have died. It will restart automatically. The kernel appears to have died. It will restart automatically.

编辑：最终的数据帧dfs_proj是一个 240 行、5 列的数据帧。

import pandas as pd
from pulp import *
from pulp import LpMaximize

dfs_proj = pd.read_csv("4for4_dfs_projections_120321.csv")
dfs_proj['count'] = 1
cols = ['Player', 'Pos', 'FFPts', 'DK ($)', 'count']
dfs_proj = dfs_proj[cols]
dfs_proj = dfs_proj[(dfs_proj['DK ($)'] >= 4000) | (dfs_proj['Pos'] == "DEF") | (dfs_proj['Pos'] == "TE")]

player_dict = dict(zip(dfs_proj['Player'], dfs_proj['count']))

# create a helper function to return the number of players assigned each position
def get_position_sum(player_vars, df, position):
    return pulp.lpSum([player_vars[i] * (position in df['Pos'].iloc[i]) for i in range(len(df))])

def get_optimals(site, data, num_lineups, optimize_on='FFPts'):
    """
    Generates x number of optimal lineups, based on the column to
    designate as the one to optimize on.
    :param str site: DK or FD. Used for salary constraints
    :param pd.DataFrame data: Pandas dataframe containing projections.
    :param int num_lineups: Number of lineups to generate.
    :param str optimize_on: Name of column in dataframe to use when optimizing
    """
    #global lineups
    lineups = []
    player_dict = dict(zip(data['Player'], data['count']))
    for i in range(1, num_lineups+1):
        prob = pulp.LpProblem('DK_NFL_weekly', pulp.const.LpMaximize)
        player_vars = []
        for row in data.itertuples():
            var = pulp.LpVariable(f'{row.Player}', cat='Binary')
            player_vars.append((row.Player, var))
        # total assigned players constraint
        prob += pulp.lpSum(player_var for player_var in player_vars) == 9
        # total salary constraint
        prob += pulp.lpSum(data['DK ($)'].iloc[i] * player_vars[i][1] for i in range(len(data))) <= 50000
        # for QB and DST, require 1 of each in the lineup
        prob += get_position_sum(player_vars, df, 'QB') == 1
        prob += get_position_sum(player_vars, df, 'DEF') == 1
        
        # to account for the FLEX position, we allow additional selections of the 3 FLEX-eligible positions: RB, WR, TE
        prob += get_position_sum(player_vars, df, 'RB') >= 2
        prob += get_position_sum(player_vars, df, 'WR') >= 3
        prob += get_position_sum(player_vars, df, 'TE') >= 1
        if i > 1:
            if optimize_on == 'Optimal Frequency':
                prob += pulp.lpSum([data['FFPts'].iloc[i] * player_vars[i][1] for i in range(len(data))]) <= (optimal - 0.001)
            else:
                prob += pulp.lpSum([data['FFPts'].iloc[i] * player_vars[i][1] for i in range(len(data))]) <= (optimal - 0.01)
        
        prob += pulp.lpSum([data['FFPts'].iloc[i] * player_vars[i][1] for i in range(len(data))])
        # solve and print the status
        prob.solve(PULP_CBC_CMD(msg=False))
        optimal = prob.objective.value()
        count = 1
        lineup = {}
        for i in range(len(data)):    
            if player_vars[i][1].value() == 1:
                row = data.iloc[i]
                lineup[f'G{count}'] = row['Player']
                count += 1
            lineup['Total Points'] = optimal
        
        lineups.append(lineup)
        players = list(lineup.values())
        for i in range(0, len(players)):
            if type(players[i]) == str:
                player_dict[players[i]] += 1
                if player_dict[players[i]] == 45:
                    data = data[data['Player'] != players[i]]
    return lineups

lineups = get_optimals(dfs_proj, 20, 'FFPts')

我尝试重新安装脚本中使用的所有库，但仍然遇到同样的问题。 即使在普通的 Python 脚本中运行它也会给我同样的错误消息。 我认为这可能与记忆有关，但我也不确定如何检查或调整。

提前感谢您的帮助！

Answer 1

您在这里有一些拼写错误...不确定您是否/如何运行它。

你有几个问题：

• 您在函数中混合了df和data变量名称。 所以谁知道那是什么。（在笔记本上工作的危险之一。）
• 在您使用player_vars的几个位置，您没有索引元组来获取变量片段，我建议您对这些使用LpVariable.dicts() ，这样更容易管理。
• 您的函数调用不考虑函数参数中的site 。

其他建议：

• 不要关闭消息传递。 您必须检查求解器输出以查看状态。 第一次尝试返回“不可行”，这就是我发现player_vars问题的方式。 如果您确实决定关闭消息，请找出一种assert(status==optimal)或冒险垃圾结果的方法。 我认为它在pulp中是可行的，我只是忘记了如何。 编辑：这里是如何。 这在使用默认的 CBC 求解器时有效，在求解后（显然）。 其他求解器，不确定：

   status = LpStatus[prob.status] assert(status=='Optimal')

• 将问题打印几次，看看它是否在构建时通过了傻笑测试。 如果你这样做了，你会看到一些施工问题。

无论如何，这对于虚假数据来说效果很好，并且可以在几秒钟内处理 20 个阵容的 1000 多名球员。

买家注意：我没有仔细审查所有约束或条件约束，所以你应该这样做。

import pandas as pd
from pulp import *
# from pulp import LpMaximize
from random import randint, choice

num_players = 1000
positions = ['RB', 'WR', 'TE', 'DEF', 'QB']
players = [(i, choice(positions), randint(1,100), randint(3000,5000), 1) for i in range(num_players)]
cols = ['Player', 'Pos', 'FFPts', 'DK ($)', 'count']
dfs_proj = pd.DataFrame.from_records(players, columns = cols)
print(dfs_proj.head())


# dfs_proj = pd.read_csv("4for4_dfs_projections_120321.csv")
# dfs_proj['count'] = 1
# cols = ['Player', 'Pos', 'FFPts', 'DK ($)', 'count']
# dfs_proj = dfs_proj[cols]

dfs_proj = dfs_proj[(dfs_proj['DK ($)'] >= 4000) | (dfs_proj['Pos'] == "DEF") | (dfs_proj['Pos'] == "TE")]

# player_dict = dict(zip(dfs_proj['Player'], dfs_proj['count']))

print(dfs_proj.head())

# create a helper function to return the number of players assigned each position
def get_position_sum(player_vars, df, position):
    return pulp.lpSum([player_vars[i][1] * (position in df['Pos'].iloc[i]) for i in range(len(df))])  #player vars not indexed

#def get_optimals(site, data, num_lineups, optimize_on='FFPts'):   # site???  # data vs df ???
def get_optimals(data, num_lineups, optimize_on='FFPts'):
    """
    Generates x number of optimal lineups, based on the column to
    designate as the one to optimize on.
    :param str site: DK or FD. Used for salary constraints
    :param pd.DataFrame data: Pandas dataframe containing projections.
    :param int num_lineups: Number of lineups to generate.
    :param str optimize_on: Name of column in dataframe to use when optimizing
    """
    #global lineups
    lineups = []
    player_dict = dict(zip(data['Player'], data['count']))
    for i in range(1, num_lineups+1):
        prob = pulp.LpProblem('DK_NFL_weekly', pulp.const.LpMaximize)
        player_vars = []
        for row in data.itertuples():
            var = pulp.LpVariable(f'P{row.Player}', cat='Binary')  # added 'P' to player name for clarity
            player_vars.append((row.Player, var))
        # total assigned players constraint
        prob += pulp.lpSum(player_var[1] for player_var in player_vars) == 9    # player var not indexed
        # total salary constraint
        prob += pulp.lpSum(data['DK ($)'].iloc[i] * player_vars[i][1] for i in range(len(data))) <= 50000
        # for QB and DST, require 1 of each in the lineup

        # !!!!  you had 'df' here which who knows what you were pulling in....  changed to data

        prob += get_position_sum(player_vars, data, 'QB') == 1
        prob += get_position_sum(player_vars, data, 'DEF') == 1
        
        # to account for the FLEX position, we allow additional selections of the 3 FLEX-eligible positions: RB, WR, TE
        prob += get_position_sum(player_vars, data, 'RB') >= 2
        prob += get_position_sum(player_vars, data, 'WR') >= 3
        prob += get_position_sum(player_vars, data, 'TE') >= 1
        if i > 1:
            if optimize_on == 'Optimal Frequency':
                prob += pulp.lpSum([data['FFPts'].iloc[i] * player_vars[i][1] for i in range(len(data))]) <= (optimal - 0.001)
            else:
                prob += pulp.lpSum([data['FFPts'].iloc[i] * player_vars[i][1] for i in range(len(data))]) <= (optimal - 0.01)
        
        prob += pulp.lpSum([data['FFPts'].iloc[i] * player_vars[i][1] for i in range(len(data))])
        print(prob)
        # solve and print the status
        prob.solve(PULP_CBC_CMD())
        optimal = prob.objective.value()
        count = 1
        lineup = {}
        for i in range(len(data)):    
            if player_vars[i][1].value() == 1:
                row = data.iloc[i]
                lineup[f'G{count}'] = row['Player']
                count += 1
            lineup['Total Points'] = optimal
        
        lineups.append(lineup)
        players = list(lineup.values())
        for i in range(0, len(players)):
            if type(players[i]) == str:
                player_dict[players[i]] += 1
                if player_dict[players[i]] == 45:
                    data = data[data['Player'] != players[i]]
    return lineups

lineups = get_optimals(dfs_proj, 10, 'FFPts')
for lineup in lineups:
    print(lineup)