使用嵌套的 defaultdict 重新分配 pandas 系列值

Question

我正在处理 NFL 数据集，并希望为 df 中的每场比赛做以下映射：

我正在尝试用每个球员到该场比赛的冲击者的距离填充一个列 (DistToRusher)。
DistToRusher 列当前填充了玩家 ID。
我正在尝试将这些播放器 ID 映射到内部字典键中的那些，并用内部字典值替换它们。
我有一个 defaultdict-of-dictionaries dist_dict 看起来像这样：

    dist_dict = {play_id1: {player_id1: distance, player_id2: distance ...}, 
                 play_id2: {player_id1: distance, player_id2: distance ...}...}

这是我的代码：

def populate_DistToRusher_column(df):
    for play_id, players_dict in dist_dict.items():
        df[df.PlayId == play_id].replace({'DistToRusher': players_dict}, inplace=True)
    return df

此代码运行缓慢（20-30 秒），并且不会更改 DistToRusher 列；当我检查 df 时，DistToRusher 仍然包含玩家 ID 号而不是距离。

这是实际数据的玩具版本：

from collections import defaultdict 
import pandas as pd
df = pd.DataFrame.from_dict({'PlayId': {
  0: 20170907000118, 1: 20170907000118, 2: 20170907000118,
  22: 20170907000139, 23: 20170907000139, 24: 20170907000139},
 'NflId': {0: 496723, 1: 2495116, 2: 2495493,
  22: 496723, 23: 2495116, 24: 2495493},
 'NflIdRusher': {0: 2543773, 1: 2543773, 2: 2543773,
  22: 2543773, 23: 2543773, 24: 2543773},
 'DistToRusher': {0: 496723, 1: 2495116, 2: 2495493,
  22: 496723, 23: 2495116, 24: 2495493}})

dist_dict = {20170907000118: defaultdict(float,
             {496723: 6.480871854928166,
              2495116: 4.593310353111358,
              2495493: 5.44898155621764}),
 20170907000139: defaultdict(float,
             {496723: 8.583355987025117,
              2495116: 5.821151088917024,
              2495493: 6.658686056573021})}

Answer 1

我认为这是对的，IIUC:

temp = pd.DataFrame(dist_dict)
df['DistToRusher2'] = df.apply(lambda x: temp[x.PlayId][x.NflId], axis=1)

or

df['DistToRusher2'] = df.apply(lambda x: dist_dict[x.PlayId][x.NflId], axis=1)

输出：

            PlayId    NflId  NflIdRusher  DistToRusher  DistToRusher2
0   20170907000118   496723      2543773        496723       6.480872
1   20170907000118  2495116      2543773       2495116       4.593310
2   20170907000118  2495493      2543773       2495493       5.448982
22  20170907000139   496723      2543773        496723       8.583356
23  20170907000139  2495116      2543773       2495116       5.821151
24  20170907000139  2495493      2543773       2495493       6.658686

Answer 2

谢谢@oppressionslayer！这就像一个魅力：

df['DistToRusher2'] = df.apply(lambda x: dist_dict[x.PlayId][x.NflId], axis=1)

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