Question

我需要创建一个DataFrame或字典。如果N = 3（其他列表中的列表数），则预期输出为：

d = {
    'xs0': [[7.0, 986.0], [17.0, 6.0], [7.0, 67.0]],
    'ys0': [[79.0, 69.0], [179.0, 169.0], [729.0, 69.0]],
    'xs1': [[17.0, 166.0], [17.0, 116.0], [17.0, 126.0]],
    'ys1': [[179.0, 169.0], [179.0, 1169.0], [1729.0, 169.0]],
    'xs2': [[27.0, 276.0], [27.0, 216.0], [27.0, 226.0]],
    'ys2': [[279.0, 269.0], [279.0, 2619.0], [2579.0, 2569.0]]
}

为此，我编写了以下代码。但我需要这段代码才能更快地运行：

import numpy as np
import pandas as pd

df_dict = {
    'X1': [1, 2, 3, 4, 5, 6, 7, 8, np.nan],
    'Y1': [9, 29, 39, 49, np.nan, 69, 79, 89, 99],
    'X2': [11, 12, 13, 14, 15, 16, 17, 18, np.nan],
    'Y2': [119, 129, 139, 149, np.nan, 169, 179, 189, 199],
    'X3': [21, 22, 23, 24, 25, 26, 27, 28, np.nan],
    'Y3': [219, 229, 239, 249, np.nan, 269, 279, 289, 299],
    'S': [123, 11, 123, 11, 123, 123, 123, 35, 123],
    'C': [9, 8, 7, 6, 5, 4, 3, 2, 1],
    'F': [1, 1, 1, 1, 2, 3, 3, 3, 3],
    'OTHER': [10, 20, 30, 40, 50, 60, 70, 80, 90],
}
bigger_df = pd.DataFrame(df_dict)

plots = [
    { 'x': 'X1', 'y': 'Y1', },
    { 'x': 'X2', 'y': 'Y2', },
    { 'x': 'X3', 'y': 'Y3', }
]

N = 3
d = {}
s_list = [123, 145, 35]
n = 0
for p in plots:
    # INITIALIZATES THE DICTIONARY ELEMENTS
    d['xs{}'.format(n)] = [[] for x in range(N)]
    d['ys{}'.format(n)] = [[] for x in range(N)]        

    # BUILDS THE LISTS FOR THOSE ELEMENTS
    for index in range(3):
        df = bigger_df.filter([p['x'], p['y'], 'S', 'F', 'C'])        # selects the minimum of columns needed
        df = df[df['F'].isin([2, 3, 4, 9]) & df[p['x']].notnull() & df[p['y']].notnull() & (df.S == s_list[index])]
        df.sort_values(['C'], ascending=[True], inplace=True)

        d['xs{}'.format(n)][index] = list(df[p['x']])
        d['ys{}'.format(n)][index] = list(df[p['y']])
    n += 1
print(d)

我想知道如果不是在循环上构建字典，我可以用pandas或numpy做一些技巧。如果结果是pandas数据帧而不是字典也对我有好处，甚至更好，但我不会，如果它会更有效率。

一些想法？

Answer 1

根据您的输入和预期输出（每个键列表中三次相同的值？），至少可以将for p in plots替换为：

for p in plots:
    # Select the data you want
    df = bigger_df.filter([p['x'], p['y'], 'S', 'F', 'C'])        # selects the minimum of columns needed
    df = df[df['F'].isin([2, 3, 4, 9]) & df[p['x']].notnull() & df[p['y']].notnull() & (df.S == 123)]   # I have used 123 to simplify, actually the value is an integer variable
    df.sort_values(['C'], ascending=[True], inplace=True)
    # fill the dictionary
    d['xs{}'.format(n)] = [list(df[p['x']]) for x in range(N)]
    d['ys{}'.format(n)] = [list(df[p['y']]) for x in range(N)]
    n += 1

至少保存for index in range(3)并在bigger_df上执行相同的操作3次。随着timeit，我用你的代码从210毫秒下降到70.5毫秒（大约三分之一）。

编辑：根据您重新定义问题的方式，我认为这可能会完成您想要的工作：

# put this code after the definition of plots
s_list = [123, 145, 35]
# create an empty DF to add your results in the loop
df_output = pd.DataFrame(index=s_list, columns=['xs0','ys0', 'xs1', 'ys1', 'xs2', 'ys2']) 
n = 0
for p in plots:
    # Select the data you want and sort them on the same line
    df_p = bigger_df[bigger_df['F'].isin([2, 3, 4, 9]) & bigger_df[p['x']].notnull() & bigger_df[p['y']].notnull() & bigger_df['S'].isin(s_list)].sort_values(['C'], ascending=[True])
    # on bigger df I would do a bit differently if the isin on F and S are the same for the three plots, 
    # I would create a df_select_FS outside of the loop before (might be faster)

    #  Now, you can do groupby on S and then you create a list of element in column p['x'] (and same for p['y'])
    # and you add them in you empty df_output in the right column
    df_output['xs{}'.format(n)] = df_p.groupby('S').apply(lambda x: list(x[p['x']]))
    df_output['ys{}'.format(n)] = df_p.groupby('S').apply(lambda x: list(x[p['y']]))
    n += 1

两个注意事项：首先，如果你的s_list中有两倍相同的值，它可能无法按你想要的方式工作，第二个条件不符合的情况（如S中的示例145 ）然后你的nan

中有df_output

如何通过从另一个更大的数据框中选择一些数据列表，以有效的方式构建一个pandas数据帧（或dict）？

1 个答案: