按赢家/输家名称查询体育数据的数据框，并获取每个球员的汇总统计数据表？

Question

我有一个网球比赛结果的 CSV 数据集，格式与此类似：https://www.kaggle.com/jordangoblet/atp-tour-20002016

我想创建另一个表格来显示每个球员的胜负数、自上次比赛以来的天数、平均发球率等。

一位朋友建议尝试这样的事情：

games = [
    ["Elliot", 7, "Maya", 10],
    ["Elliot", 10, "Brendan", 8],
    ["Jamie", 7, "Omid", 10],
    ["Elliot", 6, "Jamie", 10],
    ["Brendan", 7, "Jamie", 10],
]

players = {}
K = 32.0
game_number = 1

for game in games:
    p1name, p1score, p2name, p2score = game
    if p1name not in players:
        players[p1name] = {
            'wins': 0,
            'losses': 0,
            'score': 1000.0,
        }
    if p2name not in players:
        players[p2name] = {
            'wins': 0,
            'losses': 0,
            'score': 1000.0,
        }

更多代码...

...

print('Rank         Player         Elo       Wins    Losses')
rank = 1
for player, p in sorted(players.items(), key=lambda item: (item[1]['score'], item[0]), reverse=True):
    print('%4s %20s %7.1f %5s %7s' % (rank, player, p['score'], p['wins'], p['losses']))
    rank += 1 

我试图通过使我的 df 成为列表列表来根据我的需要调整此代码：

lol = df.values.tolist()
players = {}
for game in lol:
    game_id, tournament_name, tournament_id, tournament_date, round_match, winner, winner_id, loser, loser_id, winner_elo, 
    loser_elo, winner_delta, loser_delta, winner_set_1, loser_set_1, winner_set_2, loser_set_2, winner_set_3, loser_set_3,
    winner_set_4, loser_set_4, winner_set_5, loser_set_5, winner_sets, loser_sets, winner_serve_points_won, 
    loser_serve_points_won, winner_serve_hold, loser_serve_hold, winner_points_won, loser_points_won, winner_true_serve_pct, 
    loser_true_serve_pct = game

以上都是我原来的df中的列名

但我收到此错误： NameError: name "game_id" is not defined

原始代码运行良好，但我不明白为什么 =game 不适用于我的情况。

我错过了什么吗？有没有更好的方法来解决这个问题？我的数据集有大约 60k 个实例和 33 列

编辑：找到解决方案，变量 = 游戏应该在一行中。但是，如果有更好的解决方案来实现我在 cmets 中描述的内容，欢迎您发表评论

编辑：列信息

game/tournament/winner/loser_id = unique id's

tournament_name = matches are played in daily tournaments

tournament_date = date of tournament/match

round_match = round of tournament the match takes place

winner/loser= winner/loser name

winner/loser_elo = current elo for the day

winner/loser_delta = delta change after match

winner/loser_set_1,2,3,4,5 = score for winner/loser per game, match ends when a player reaches 3 games, 11 points needed to win a game or 2 point differential (extra points if game is tied at 10-10)

winner/loser_sets= number of sets

winner/loser_points_won = total points per player in the match

winner/loser_true_serve_pct = % of points player wins when he is serving

winner_serve_points_won/loser_serve_points_won is obsolete

winner_serve_hold/loser_serve_hold will be replaced by winner/loser_true_serve_pct

Answer 1

您需要查询出现在“赢家”或“输家”列中的球员姓名的数据框，以获取matches，这是一个包含该球员的所有比赛的数据框。这比乍一看更难 - 我们不能只使用简单的df.groupby()。抱歉低估了您的问题。

• 以下解决方案使用 df[['Winner','Loser']].isin([player]).any(axis=1)

  • 请注意，参数化名称可能出现的列列表，它不会对它们进行硬编码。比df['Winner'] == player or df['Loser'] == player更优雅的成语
  • 我还研究了df.query("...")，它接受任意 SQL 样式的查询字符串，但它的语法很烦人

• 两列共有 1387 名独立玩家：df['Winner'] 中的 879 名独立玩家和 df['Loser'] 中的 1383 名。我们将它们组合成一组player_names。您可以使用列表，但使用 set 删除重复项更容易：set(df['Winner'].unique()) | set(df['Loser'].unique()) where '|'是 setwise-'or'/.union。

• 为了提高处理玩家姓名的效率，我们使用分类而不是字符串。请注意代码必须如何在“赢家”、“输家”列中拥有一个具有合并类别值的统一分类（参见代码）。
• 我们将defaultdict(dict) 用于players，这很优雅，因为现在您可以直接定义字段players[player]['WhateverField']，无论是否定义了字典players[player]，否则将自动创建。
• 此处的代码显示了您要创建的聚合字段的示例。
  • 注意如何直接使用 pandas 矢量化 mean()、sum() 等来处理简单的事情
  • 要创建其他更任意的聚合，您可以使用matches.agg(...) 和内部的自定义/lambda 函数。

代码：

import pandas as pd

# https://www.kaggle.com/jordangoblet/atp-tour-20002016
df = pd.read_csv('../input/Data.csv', encoding='latin1', usecols=range(12+1), na_values='NR',
    parse_dates=['Date'], dayfirst=True)
# make sure that the 'NR' in WRank/LRank column is recognized as NaN, so they read in as float columns
# Columns 0-12 (basic data), 13-25 (results by set), 26-53 (betting odds: float)

# Create table of number of wins, losses, days since last match, average serve percentage etc. per player.

pd.options.display.max_rows = 100
pd.options.display.precision = 3    

player_names = set(df['Winner'].unique()) | set(df['Loser'].unique())
player_names = sorted(player_names) # 1.0.x BUG: order seems totally unstable(!)
player_dtype = pd.CategoricalDtype(categories = player_names) # merge across multiple columns

# WRONG: categories only taken from each individual column, not merged
# df[['WinnerCat1','LoserCat1']] = df[['Winner','Loser']].astype('category')
# RIGHT:
df[['WinnerCat','LoserCat']] = df[['Winner','Loser']].astype(player_dtype)

from collections import defaultdict
players = defaultdict(dict)

for _,player in (player_names):
    # query df where player occurs in either column, can't use `df[['Winner','Loser']].groupby(player)`
    matches = df[ df[['Winner','Loser']].isin([player]).any(axis=1) ]

    print(f'\n[MATCHES FOR {_} {player}]')
    print(matches.iloc[:, 0:13])

    players[player]['Wins']   = matches['Winner'].eq(player).sum()
    players[player]['Losses'] = matches['Loser'].eq(player).sum()
    players[player]['Avg']    = matches['Winner'].eq(player).mean().round(4)
    # You can do any arbitrary calculation involving both players' columns, e.g....
    players[player]['RankDiff'] = (matches['WRank'] - matches['LRank']).mean()

    #if _ >= 2: break # for debugging

Answer 2

例如，如果你有一个 csv 文件，就这样加载它

import pandas as pd
dataframe = pd.read.csv('file.csv')

现在，选择您感兴趣的列（我将使用 kaggle 数据库作为参考）。

请注意，我不会写下您提出的所有问题，因为我没有完全理解您的问题（kaggle 数据库上没有日期，以及每个玩家的平均时间？）

dataframe = dataframe['Winner', 'Loser']

现在，我们使用 groupby 函数和 count 函数来获取每个玩家的输赢数量（同时使其在“赢家”/“输家”列上的出现独一无二。

dataframe['Wins'] = dataframe.groupby(labels='Winner', axis=1).count()
dataframe['Losses'] = dataframe.groupby(labels='Loser', axis=1).count()

Answer 3

我运行以下代码没有任何错误，所以我认为 = 游戏没有问题。

df = pd.read_csv(r'Ten.csv')


lol = df.values.tolist()


players = {}
for game in lol:
    Player,Tournament,Wins,Losses = game

    if Player not in players:
        players[Player] = {
            'Tournament': 'foo',
            'wins': 1,
            'losses': 2,
        }

print(players)

但是如果 Player 不同，例如以下（小写“p”）

player,Tournament,Wins,Losses = game

或：（[播放器]末尾的“s”）

        players[Players] = {
            'Tournament': 'foo',
            'wins': 1,
            'losses': 2,
        }

会弹出错误 NameError: name 'Player' is not defined。您是否 100% 确定 game_id 没有拼错？