查找所有连胜最长的玩家

Question

我正在尝试解决使用 Python 的3.2. 找到最长连胜纪录的玩家的问题.谁能帮我一个更好的解决方案？另外，我很好奇如何调整这样的解决方案来输出每月或每年最长连胜的球员？

下面是示例数据框players_results

Player_id     Match_Date             Match_Result
401           2021-05-04 00:00:00    W
401           2021-05-09 00:00:00    L
401           2021-05-16 00:00:00    W
401           2021-05-18 00:00:00    W
401           2021-05-22 00:00:00    L
401           2021-06-15 00:00:00    L
401           2021-06-16 00:00:00    W
401           2021-06-18 00:00:00    W
402           2021-05-14 00:00:00    L
402           2021-05-23 00:00:00    L
402           2021-05-24 00:00:00    W
402           2021-06-01 00:00:00    W
402           2021-06-02 00:00:00    W
402           2021-07-01 00:00:00    W
403           2021-05-03 00:00:00    L
403           2021-05-11 00:00:00    W
403           2021-05-12 00:00:00    W
403           2021-05-13 00:00:00    W
403           2021-05-20 00:00:00    W
403           2021-05-25 00:00:00    W
403           2021-07-06 00:00:00    L
404           2021-05-10 00:00:00    W
404           2021-05-16 00:00:00    W
404           2021-05-20 00:00:00    W
404           2021-05-22 00:00:00    W
404           2021-05-28 00:00:00    L
405           2021-05-07 00:00:00    L
405           2021-05-25 00:00:00    W
405           2021-06-06 00:00:00    L
405           2021-06-07 00:00:00    W
405           2021-06-14 00:00:00    W
405           2021-07-01 00:00:00    W

预期输出

player_id     longest_winningstreak
403           5

我的不可扩展代码

df = players_results.groupby('player_id').count()
df = df['match_date'].reset_index()
#record result of player_id = 401 into vector b
a = [] # record the number of consecutive "W" of player_id = 401
x = 0
for i in range(df['match_date'][0]):
    if players_results['match_result'][i] == 'W':
        x = x + 1
        if (i == df['match_date'][0]-1):
            a.append(x)
    else:
        a.append(x)
        x = 0
print(a)
b = []
b.append([df['player_id'][0],max(a)])
print(b)

c = []
y=0
for i in range(df['match_date'][0], df['match_date'][0]+df['match_date'][1]):
    if players_results['match_result'][i] == 'W':
        y = y + 1
        if (i == df['match_date'][0]+df['match_date'][1]-1):
            c.append(y)
    else:
        c.append(y)
        y = 0
#record result of player_id = 402 into the vector b
b.append([df['player_id'][1],max(c)])

d = []
z=0
for i in range(df['match_date'][0]+df['match_date'][1], df['match_date'] [0]+df['match_date'][1]+df['match_date'][2]):
    if players_results['match_result'][i] == 'W':
        z = z + 1
        if (i == df['match_date'][0]+df['match_date'][1]+df['match_date'][2]-1):
            d.append(z)
    else:
        d.append(z)
        z = 0
b.append([df['player_id'][2],max(d)])

e = []
z2=0
for i in range(df['match_date'][0]+df['match_date'][1]+df['match_date'][2], df['match_date'][0]+df['match_date'][1]+df['match_date'][2]+df['match_date'][3]):
    if players_results['match_result'][i] == 'W':
        z2 = z2 + 1
        if (i == df['match_date'][0]+df['match_date'][1]+df['match_date'][2]+df['match_date'][3]-1):
            e.append(z2)
    else:
        e.append(z2)
        z2 = 0
#print(e)
b.append([df['player_id'][3],max(e)])

f = []
z3=0
for i in range(df['match_date'][0]+df['match_date'][1]+df['match_date'][2]+df['match_date'][3], df['match_date'][0]+df['match_date'][1]+df['match_date'][2]+df['match_date'][3]+df['match_date'][4]):
    if players_results['match_result'][i] == 'W':
        z3 = z3 + 1
        if (i == df['match_date'][0]+df['match_date'][1]+df['match_date'][2]+df['match_date'][3]+df['match_date'][4]-1):
            f.append(z3)
    else:
        f.append(z3)
        z3 = 0
#print(e)
b.append([df['player_id'][4],max(f)])   

Answer 1

过滤由累积和创建的组，比较不等于W 和SeriesGroupBy.value_counts，然后通过Series.agg 和Series.idxmax 和max 使用player_id 获得最大值：

m = df['Match_Result'].str.upper().ne('W')

s = m.cumsum()[~m].groupby(df['Player_id']).value_counts().reset_index(level=1, drop=True)

df = s.agg({'player_id': 'idxmax', 'longest_winningstreak':'max'}).to_frame(0).T
print (df)
   player_id  longest_winningstreak
0        403                      5

每月解决方案：

df['Match_Date'] = pd.to_datetime(df['Match_Date'])

m = df['Match_Result'].ne('W')

s = (m.cumsum()[~m].groupby([df['Player_id'], df['Match_Date'].dt.to_period('m')])
       .value_counts()
       .reset_index(level=-1, drop=True))

df1 = (s.groupby(level=0)
        .agg([('period',lambda x: x.idxmax()[1]),('longest_winningstreak','max')]))

print (df1)
            period  longest_winningstreak
Player_id                                
401        2021-06                      2
402        2021-06                      2
403        2021-05                      5
404        2021-05                      4
405        2021-06                      2

Answer 2

一种使用itertools.groupby的方式：

from itertools import groupby

s = df["Match_Result"].str.lower().eq("w")

def longest_pattern(ser):
    lens = [len(list(g)) for k, g in groupby(ser) if k]
    return max(lens)

new_df = s.groupby(df["Player_id"]).apply(longest_pattern)

或者有点诡计，但使用re.findall的另一种方式：

import re

def longest(string):
    return max(len(i) for i in re.findall("w+", string, flags=re.I))

new_df = df.groupby("Player_id")["Match_Result"].sum().apply(longest)

然后您可以使用pandas.Series.nlargest 获得所需的输出：

new_df.nlargest(1)

输出：

Player_id
403    5
Name: Match_Result, dtype: int64