在Pandas中最快的计算方法？ [英] Fastest way to calculate in Pandas?

问题描述

给出以下两个数据框：

df1 =
     Name  Start  End
  0  A     10     20
  1  B     20     30
  2  C     30     40

df2 =
     0   1
  0  5   10
  1  15  20
  2  25  30

df2 没有列名，但您可以假设第0列的偏移量是 df1.Start ，而第1列的偏移量是 df1.End 。我想将 df2 换位到 df1 上以获得开始和结束差异。最终的 df1 数据帧应如下所示：

df2 has no column names, but you can assume column 0 is an offset of df1.Start and column 1 is an offset of df1.End. I would like to transpose df2 onto df1 to get the Start and End differences. The final df1 dataframe should look like this:

  Name  Start  End  Start_Diff_0  End_Diff_0  Start_Diff_1  End_Diff_1  Start_Diff_2  End_Diff_2
0    A     10   20             5          10            -5           0           -15         -10
1    B     20   30            15          20             5          10            -5           0
2    C     30   40            25          30            15          20             5          10

我有一个可行的解决方案，但我不满意，因为它花费的时间太长在处理具有数百万行的数据框时运行。下面是一个示例测试案例，用于模拟处理30,000行。可以想象，在1GB数据帧上运行原始解决方案（method_1）将是一个问题。是否可以使用Pandas，Numpy或其他软件包来更快地完成此操作？

I have a solution that works, but I'm not satisfied with it because it takes too long to run when processing a dataframe that has millions of rows. Below is a sample test case to simulate processing 30,000 rows. As you can imagine, running the original solution (method_1) on a 1GB dataframe is going to be a problem. Is there a faster way to do this using Pandas, Numpy, or maybe another package?

更新：我已将提供的解决方案添加到

UPDATE: I've added the provided solutions to the benchmarks.

# Import required modules
import numpy as np
import pandas as pd
import timeit

# Original
def method_1():
    df1 = pd.DataFrame([['A', 10, 20], ['B', 20, 30], ['C', 30, 40]] * 10000, columns=['Name', 'Start', 'End'])
    df2 = pd.DataFrame([[5, 10], [15, 20], [25, 30]], columns=None)
    # Store data for new columns in a dictionary
    new_columns = {}
    for index1, row1 in df1.iterrows():
        for index2, row2 in df2.iterrows():
            key_start = 'Start_Diff_' + str(index2)
            key_end = 'End_Diff_' + str(index2)
            if (key_start in new_columns):
                new_columns[key_start].append(row1[1]-row2[0])
            else:
                new_columns[key_start] = [row1[1]-row2[0]]
            if (key_end in new_columns):
                new_columns[key_end].append(row1[2]-row2[1])
            else:
                new_columns[key_end] = [row1[2]-row2[1]]
    # Add dictionary data as new columns
    for key, value in new_columns.items():
        df1[key] = value

# jezrael - https://stackoverflow.com/a/60843750/452587
def method_2():
    df1 = pd.DataFrame([['A', 10, 20], ['B', 20, 30], ['C', 30, 40]] * 10000, columns=['Name', 'Start', 'End'])
    df2 = pd.DataFrame([[5, 10], [15, 20], [25, 30]], columns=None)
    # Convert selected columns to 2d numpy array
    a = df1[['Start', 'End']].to_numpy()
    b = df2[[0, 1]].to_numpy()
    # Output is 3d array; convert it to 2d array
    c = (a - b[:, None]).swapaxes(0, 1).reshape(a.shape[0], -1)
    # Generate columns names and with DataFrame.join; add to original
    cols = [item for x in range(b.shape[0]) for item in (f'Start_Diff_{x}', f'End_Diff_{x}')]
    df1 = df1.join(pd.DataFrame(c, columns=cols, index=df1.index))

# sammywemmy - https://stackoverflow.com/a/60844078/452587
def method_3():
    df1 = pd.DataFrame([['A', 10, 20], ['B', 20, 30], ['C', 30, 40]] * 10000, columns=['Name', 'Start', 'End'])
    df2 = pd.DataFrame([[5, 10], [15, 20], [25, 30]], columns=None)
    # Create numpy arrays of df1 and df2
    df1_start = df1.loc[:, 'Start'].to_numpy()
    df1_end = df1.loc[:, 'End'].to_numpy()
    df2_start = df2[0].to_numpy()
    df2_end = df2[1].to_numpy()
    # Use np tile to create shapes that allow elementwise subtraction
    tiled_start = np.tile(df1_start, (len(df2), 1)).T
    tiled_end = np.tile(df1_end, (len(df2), 1)).T
    # Subtract df2 from df1
    start = np.subtract(tiled_start, df2_start)
    end = np.subtract(tiled_end, df2_end)
    # Create columns for start and end
    start_columns = [f'Start_Diff_{num}' for num in range(len(df2))]
    end_columns = [f'End_Diff_{num}' for num in range(len(df2))]
    # Create dataframes of start and end
    start_df = pd.DataFrame(start, columns=start_columns)
    end_df = pd.DataFrame(end, columns=end_columns)
    # Lump start and end into one dataframe
    lump = pd.concat([start_df, end_df], axis=1)
    # Sort the columns by the digits at the end
    filtered = lump.columns[lump.columns.str.contains('\d')]
    cols = sorted(filtered, key=lambda x: x[-1])
    lump = lump.reindex(cols, axis='columns')
    # Hook lump back to df1
    df1 = pd.concat([df1,lump],axis=1)

print('Method 1:', timeit.timeit(method_1, number=3))
print('Method 2:', timeit.timeit(method_2, number=3))
print('Method 3:', timeit.timeit(method_3, number=3))

输出：

Method 1: 50.506279182
Method 2: 0.08886280600000163
Method 3: 0.10297686199999845

推荐答案

我建议在此处使用numpy-将选定的列转换为 2d numpy 数组：

I suggest use here numpy - convert selected columns to 2d numpy array in first step::

a = df1[['Start','End']].to_numpy()
b = df2[[0,1]].to_numpy()

输出为3d数组，将其转换为 2d数组：

Output is 3d array, convert it to 2d array:

c = (a - b[:, None]).swapaxes(0,1).reshape(a.shape[0],-1)
print (c)
[[  5  10  -5   0 -15 -10]
 [ 15  20   5  10  -5   0]
 [ 25  30  15  20   5  10]]

最后生成列名称，并使用 DataFrame.join 添加到原始内容：

Last generate columns names and with DataFrame.join add to original:

cols = [item for x in range(b.shape[0]) for item in (f'Start_Diff_{x}', f'End_Diff_{x}')]
df = df1.join(pd.DataFrame(c, columns=cols, index=df1.index))
print (df)
  Name  Start  End  Start_Diff_0  End_Diff_0  Start_Diff_1  End_Diff_1  \
0    A     10   20             5          10            -5           0   
1    B     20   30            15          20             5          10   
2    C     30   40            25          30            15          20   

   Start_Diff_2  End_Diff_2  
0           -15         -10  
1            -5           0  
2             5          10  

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