我有一个包含3列的数据集-DriverID，Race，Place。驱动程序ID
DriverID Race Place
83 1 1
18 1 2
20 1 3
48 1 4
53 1 5
对于每场比赛，我想计算一个{strong> Place的{{1}}列中的成对差异的矩阵（numpy数组）。问题在于，并非所有DriverID和种族都被代表。因此，我决定首先为DriverID和DriverID的每个组合创建一个完整的交叉联接表，如下所示（下面的可复制示例）：
Race现在要获取成对差异，我按如下进行操作（使用here中的方法：
url = "http://personal.psu.edu/drh20/code/btmatlab/nascar2002.txt"
races_trimmed = pd.read_table(url, sep=" ")
# Create a cartesian product of unique drivers and races to get every combination
unq_drivers = sorted(races_trimmed["DriverID"].unique())
unq_drivers = [x for x in unq_drivers if str(x) != 'nan']
unq_races = sorted(races_trimmed["Race"].unique())
unq_races = [x for x in unq_races if str(x) != 'nan']
# Get a dataframe
unq_drivers_df = pd.DataFrame(unq_drivers, columns=["DriverID"])
unq_races_df = pd.DataFrame(unq_races, columns=["Race"])
# Let's cross join the columns to get all unique combinations of drivers and races
all_driver_race_combs = unq_drivers_df.assign(foo=1).merge(unq_races_df.assign(foo=1)).drop('foo', 1)
all_driver_race_combs = all_driver_race_combs.sort_values(by=['Race', 'DriverID'])
all_driver_race_mg = pd.merge(all_driver_race_combs, races_trimmed, how='left',
left_on=['DriverID','Race'], right_on = ['DriverID','Race'])
您会看到，它输出# Now let's do a pairwise difference in finish across drivers for a
# single race
# based on https://stackoverflow.com/questions/46266633/pandas-creating-difference-matrix-from-data-frame
race_num = 2.0
race_res = all_driver_race_mg[all_driver_race_mg["Race"] == race_num]
race_res = race_res.sort_values(by=['DriverID'])
arr = (race_res['Place'].values - race_res['Place'].values[:, None])
new_race_1 = pd.concat((race_res['DriverID'], pd.DataFrame(arr, columns=race_res['DriverID'])), axis=1)
# Remove the first column - it has the DriverID in the pairwise matrix
new_race_1 = new_race_1.values[:, 1:]
new_race_1.shape
数组，而不是(166, 83)的{{1}}数组。对于(83, 83)，它起作用，但对于所有其他种族，它不起作用。谁能解释如何校正计算，即为每个有效的race_num = 2.0输出race_num = 1.0矩阵？我认为这是83 * 83值，但不确定如何解决？