python - Pandas / NumPy : Fastest way to create a ladder?

Question

我有一个像这样的 Pandas 数据框:

    color     cost    temp
0   blue      12.0    80.4   
1    red       8.1    81.2 
2   pink      24.5    83.5

我想为每一行创建一个成本“阶梯”或“范围”，增量为 50 美分，从低于当前成本 0.50 美元到高于当前成本 0.50 美元。我当前的代码类似于以下内容:

incremented_prices = []

df['original_idx'] = df.index # To know it's original label

for row in df.iterrows():
    current_price = row['cost']
    more_costs    = numpy.arange(current_price-1, current_price+1, step=0.5)

    for cost in more_costs:
        row_c = row.copy()
        row_c['cost'] = cost
        incremented_prices.append(row_c)

df_incremented = pandas.concat(incremented_prices)

这段代码将生成一个 DataFrame，如下所示:

    color     cost    temp  original_idx
0   blue      11.5    80.4            0
1   blue      12.0    80.4            0 
2   blue      12.5    80.4            0  
3    red       7.6    81.2            1 
4    red       8.1    81.2            1 
5    red       8.6    81.2            1 
6   pink      24.0    83.5            2
7   pink      24.5    83.5            2
8   pink      25.0    83.5            2

在真正的问题中，我将范围从 -$50.00 到 $50.00，我发现这真的很慢，有没有更快的向量化方法？

Answer 1

您可以尝试使用 numpy.repeat 重新创建数据框:

cost_steps = pd.np.arange(-0.5, 0.51, 0.5)
repeats = cost_steps.size   

pd.DataFrame(dict(
    color = pd.np.repeat(df.color.values, repeats),
    # here is a vectorized method to calculate the costs with all steps added with broadcasting
    cost = (df.cost.values[:, None] + cost_steps).ravel(),
    temp = pd.np.repeat(df.temp.values, repeats),
    original_idx = pd.np.repeat(df.index.values, repeats)
    ))

更新更多列:

df1 = df.rename_axis("original_idx").reset_index()
cost_steps = pd.np.arange(-0.5, 0.51, 0.5)
repeats = cost_steps.size   

pd.DataFrame(pd.np.hstack((pd.np.repeat(df1.drop("cost", 1).values, repeats, axis=0),
                          (df1.cost[:, None] + cost_steps).reshape(-1, 1))),
             columns=df1.columns.drop("cost").tolist()+["cost"])