Python中将列表元素嵌套到数据框

Question

公平警告这个问题确实需要一个非标准的 Python 包，nba_api。我有一个包含 3 个元素的列表，列表中的每个元素都包含另一个包含 2 个元素的列表：player 数据框和 team 数据框。实现以下预期结果的推荐方法是：1 个组合 player 数据框和 1 个组合 team 数据框？来自 R 背景，我将通过以下方式解决此问题：1. 将 players 数据框与 team 数据框连接到 joined_list 然后，2. 使用 do.call(rbind, joined_list) 将结果行绑定到一个数据框.我知道这对于许多有经验的 Python 用户来说可能是非常初级的，但是在经过多次搜索后，我正在努力寻找正确的方法。

import nba_api
import requests
import pandas as pd

from nba_api.stats.endpoints import boxscoreadvancedv2

# vector of game ids (test purposes)
gameids = ['0021900001','0021900002','0021900012']

headers1 = {
    'Host': 'stats.nba.com',
    'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:61.0) Gecko/20100101 Firefox/61.0',
    'Accept': 'application/json, text/plain, */*',
    'Accept-Language': 'en-US,en;q=0.5',
    'Referer': 'https://stats.nba.com/',
    'Accept-Encoding': 'gzip, deflate, br',
    'Connection': 'keep-alive',
}

# store player and team results for each gameids as elements of list temp
temp = list()
for i in range(len(gameids)):
    temp.append(boxscoreadvancedv2.BoxScoreAdvancedV2(game_id = gameids[i], headers=headers1))

# manually access elements of list and output to data frame
## there has to be an easier way to access list elements and rowbind the results!!!
df_out0 = temp[0].get_data_frames()
df_player0 = df_out0[0]
df_team0 = df_out0[1]

df_out1 = temp[1].get_data_frames()
df_player1 = df_out1[0]
df_team1 = df_out1[1]

Answer 1

经过更多阅读（和清晰）后，我能够将代码的手动部分组合到 for 循环中，生成一个包含球员数据的列表和一个包含团队数据的列表。然后，使用这篇文章：Concatenate a list of pandas dataframes together 我能够将player 和team 列表合并到各自的数据框中。

## output player frames
i=0
df_out=[]
df_players=[]
for i in range(len(temp)):
    df_out = temp[i].get_data_frames()
    df_players.append(df_out[0])         # index 0 will always contain player frame

df_players = pd.concat(df_players)
print(df_players)

## output team frames
i=0
df_out=[]
df_team=[]
for i in range(len(temp)):
    df_out = temp[i].get_data_frames()
    df_team.append(df_out[1])            # index 1 will always contain team frame

df_team = pd.concat(df_team)
print(df_team)

Answer 2

首先，恭喜您坚持并自己找到了解决方案！ :D

评论和提示

你可以直接遍历一个列表，不需要索引

lst_1 = [1, 2, 3, 4]

for i in range(len(lst_1)):
    print(i)

可以写成

lst_1 = [1, 2, 3, 4]

for item in lst_1:
    print(item)

List comprehensions 和 generator expressions 太棒了

奖励：注意我对变量名所做的更改。有关 Python 样式的一般参考，请参阅 PEP 8。

gameids = ['0021900001','0021900002','0021900012']

headers1 = {
    'Host': 'stats.nba.com',
    'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:61.0) Gecko/20100101 Firefox/61.0',
    'Accept': 'application/json, text/plain, */*',
    'Accept-Language': 'en-US,en;q=0.5',
    'Referer': 'https://stats.nba.com/',
    'Accept-Encoding': 'gzip, deflate, br',
    'Connection': 'keep-alive',
}

# store player and team results for each gameids as elements of list temp
temp = list()
for i in range(len(gameids)):
    temp.append(boxscoreadvancedv2.BoxScoreAdvancedV2(game_id = gameids[i], headers=headers1))

可以写成

game_ids = ['0021900001','0021900002','0021900012']

api_headers = {
    'Host': 'stats.nba.com',
    'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:61.0) Gecko/20100101 Firefox/61.0',
    'Accept': 'application/json, text/plain, */*',
    'Accept-Language': 'en-US,en;q=0.5',
    'Referer': 'https://stats.nba.com/',
    'Accept-Encoding': 'gzip, deflate, br',
    'Connection': 'keep-alive',
}

api_results = [boxscoreadvancedv2.BoxScoreAdvancedV2(game_id=curr_game_id, headers=api_headers) for curr_game_id in game_ids]

你在同一件事上迭代了两次

# output player frames
i=0
df_out=[]
df_players=[]
for i in range(len(temp)):
    df_out = temp[i].get_data_frames()
    df_players.append(df_out[0])         # index 0 will always contain player frame

df_players = pd.concat(df_players)
print(df_players)

# output team frames
i=0
df_out=[]
df_team=[]
for i in range(len(temp)):
    df_out = temp[i].get_data_frames()
    df_team.append(df_out[1])            # index 1 will always contain team frame

df_team = pd.concat(df_team)
print(df_team)

使用前两个技巧，我们最终得到的结果如下：

players_lst = []
team_lst = []

for curr_res in api_results:
    curr_dfs = curr_res.get_data_frames()
    players_lst.append(curr_dfs[0])
    team_lst.append(curr_dfs[1])

players_df = pd.concat(players_lst)
team_df = pd.concat(team_lst)

我的解决方案

在这里，为了清楚起见，稍微细分一下。

import pandas as pd
from nba_api.stats.endpoints.boxscoreadvancedv2 import BoxScoreAdvancedV2

game_ids = ['0021900001', '0021900002', '0021900012']

api_headers = {
    'Host': 'stats.nba.com',
    'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:61.0) Gecko/20100101 Firefox/61.0',
    'Accept': 'application/json, text/plain, */*',
    'Accept-Language': 'en-US,en;q=0.5',
    'Referer': 'https://stats.nba.com/',
    'Accept-Encoding': 'gzip, deflate, br',
    'Connection': 'keep-alive',
}

# generator of results from the API
api_results = (BoxScoreAdvancedV2(game_id=curr_game_id, headers=api_headers) for curr_game_id in game_ids)

# generator of lists of DataFrames from the API results
# think of it like: [[Player DF, Team DF], [Player DF, Team DF], ...]
api_res_dfs = (curr_res.get_data_frames() for curr_res in api_results)

# unpacking the size 2 lists of DataFrames into 2 flat lists
# [[Player DF, Team DF], [Player DF, Team DF], ...] -> [Player DF, Player DF, ...], [Team DF, Team DF, ...]
# see https://stackoverflow.com/q/2921847/11301900 for more on the use of the asterisk (*)
players_tupe, team_tupe = zip(*api_res_dfs)

# concatenating the various DataFrames, exactly the same as in your original code
players_df = pd.concat(players_tupe)
team_df = pd.concat(team_tupe)

print(players_df)
print(team_df)

这取决于这样一个事实，正如您所指出的，玩家 DataFrame 始终位于列表中的第一位，而团队 DataFrame 始终位于第二位，而且这些是列表中的唯一两项结果列表。

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如果您有任何问题，请告诉我 :)