如何循环并计算每场比赛已经玩了多少次？答案

【问题标题】：How to loop and count how many times each game has been played?如何循环并计算每场比赛已经玩了多少次？
【发布时间】：2021-12-13 05:16:33
【问题描述】：

game_record = [['*', 'G02', 'G05', 'G07', 'G08', 'G10'], 
          ['P001', '1', '0', '-1', '503', '1'], 
          ['P067', '1', '1', '0', '-1', '503'], 
          ['P218', '0', '1', '1', '-1', '-1'], 
          ['P101', '0', '0', '1', '1', '503'], 
          ['P456', '1', '1', '-1', '1', '-1']]

game_record 列表数据的信息：

* = 没有，只是一个符号
GXX = game_id
PXXX = player_id
1 = 玩游戏并获胜（游戏结束并记录）
0 = 玩游戏但输了（游戏结束并记录）
-1 = 不玩游戏（未录制）
503 = 在正在进行的游戏中（游戏未完成且未录制）

例如：

G02 已由 5 名玩家完成，共 3 名获胜和 2 名失败。
G08 已由 2 名玩家完成并完成，共有 2 名玩家获胜。

因此只有“1”和“0”才有意义。

我想循环这个 2D 列表来计算每个游戏 (game_id) 被玩家玩了多少次、赢了和输了多少次。然后将它们存储到字典中。

预期输出：

Game_info = {G02: [5, 3, 2], G05: [5, 3, 2], G07: [3, 2, 1], G08: [2, 2, 0], G10: [1, 1, 0]}

【问题讨论】：

标签： python python-3.x loops dictionary

【解决方案1】：

从字典开始，game_record 中的所有游戏都用[0,0,0] 初始化

Game_info = { game:[0,0,0] for game in game_record[0] if game != "*"}

然后您可以遍历game_record 并更新Game_info 中的值

n, m = len(game_record[0]), len(game_record)

for y in range(1,n):
     g_id = game_record[0][y]
     for x in range(1,m):
         res = game_record[x][y]
         if res == '1':
             Game_info[g_id][0] += 1
             Game_info[g_id][1] += 1
         elif res == '0':
             Game_info[g_id][0] += 1
             Game_info[g_id][2] += 1

【讨论】：

【解决方案2】：

您可以使用NumPy 和布尔数组来保存循环：

import numpy as np

results = np.array(game_record)[1:, 1:].astype(int)

won_by_game = (results == 1).sum(axis=0)
lost_by_game = (results == 0).sum(axis=0)
played_by_game = won_by_game + lost_by_game

game_info = {game: [played, won, lost] 
             for game, played, won, lost in 
             zip(game_record[0][1:], played_by_game, 
                 won_by_game, lost_by_game)}
game_info

{'G02': [5, 3, 2],
 'G05': [5, 3, 2],
 'G07': [3, 2, 1],
 'G08': [2, 2, 0],
 'G10': [1, 1, 0]}

【讨论】：

【解决方案3】：

试试这个。

game_record = [
    ['*', 'G02', 'G05', 'G07', 'G08', 'G10'], 
    ['P001', '1', '0', '-1', '503', '1'], 
    ['P067', '1', '1', '0', '-1', '503'], 
    ['P218', '0', '1', '1', '-1', '-1'], 
    ['P101', '0', '0', '1', '1', '503'], 
    ['P456', '1', '1', '-1', '1', '-1']
]


res = {}

for i in range(1, 6):
    w, l = 0, 0
    label = game_record[0][i]
    for j in range(1, 6):   
        v = game_record[j][i]  # access matrix value (column-wise) except row 0 and col 0
        if v == '-1' or v == '503':
            continue
        if v == '1':
            w += 1
        else:
            l += 1
    g = w + l
    res.update({label: [g, w, l]})

print(res)
# {'G02': [5, 3, 2], 'G05': [5, 3, 2], 'G07': [3, 2, 1], 'G08': [2, 2, 0], 'G10': [1, 1, 0]}

【讨论】：

【解决方案4】：

这是我没有库的解决方案。

names = game_record[0][1:]  
game_record = [*zip(*game_record[1:])][1:]  # Transpose and drop
game_record = [list(map(int, x)) for x in game_record]  # Turn into int
solution = dict()
for i, game in enumerate(game_record):
    game = [x for x in game if x not in [-1, 503]] # Filter unwanted
    solution[names[i]] = [len(game), sum(game), len(game) - sum(game)]

【讨论】：