如何使用基于python中的行的计算列表来扩展数据框答案

【问题标题】：How to expand dataframe with a calculated list based on rows in python如何使用基于python中的行的计算列表来扩展数据框
【发布时间】：2022-08-14 00:05:22
【问题描述】：

我有这段代码生成的那种数据

import pandas as pd

def multichoose(n,k):
    if k < 0 or n < 0: return \"Error\"
    if not k: return [[0]*n]
    if not n: return []
    if n == 1: return [[k]]
    return [[0]+val for val in multichoose(n-1,k)] + \\
        [[val[0]+1]+val[1:] for val in multichoose(n,k-1)]

states=[]
for i in range(0,3):
    states=states+multichoose(3,i)
df_states = pd.DataFrame(states,columns=[\'x1\',\'x2\',\'x3\'])
df_states[\'cumsum\']=df_states[\'x1\']+df_states[\'x2\']+df_states[\'x3\']

x1  x2  x3  cumsum
0   0   0   0
0   0   1   1
0   1   0   1
1   0   0   1
0   0   2   2
0   1   1   2
0   2   0   2
1   0   1   2
1   1   0   2
2   0   0   2

我想用这个计算来扩展我的数据

# For example for the first row
# in range of cumsum value + 2
[[x, y] for x in range(df_states[[\'cumsum\']].iloc[0][0]+2) for y in range(df_states[[\'cumsum\']].iloc[0][0]+2)]
    
#output
[[0, 0], [0, 1], [1, 0], [1, 1]]

所以我的预期结果是


x1  x2  x3  cumsum a1 a2
0   0   0   0      0  0
0   0   0   0      0  1
0   0   0   0      1  0
0   0   0   0      1  1
0   0   1   1      0  0
0   0   1   1      0  1
0   0   1   1      0  2
0   0   1   1      1  0
0   0   1   1      1  1
0   0   1   1      1  2
0   0   1   1      2  0
0   0   1   1      2  1
0   0   1   1      2  2
.   .   .   .      .  .
.   .   .   .      .  .

在最终结果中，我需要为所有行实现此扩展

感谢您的帮助<3

标签： python pandas expand

【解决方案1】：

亲爱的trying_to_be_a_dev<

我希望你一切都好

import pandas as pd


def multichoose(n, k):
    if k < 0 or n < 0: return "Error"
    if not k: return [[0] * n]
    if not n: return []
    if n == 1: return [[k]]
    return [[0] + val for val in multichoose(n - 1, k)] + \
           [[val[0] + 1] + val[1:] for val in multichoose(n, k - 1)]


states = []
for i in range(0, 3):
    states = states + multichoose(3, i)
df_states = pd.DataFrame(states, columns=['x1', 'x2', 'x3'])
df_states['cumsum'] = df_states['x1'] + df_states['x2'] + df_states['x3']
df_states['a1'] = ''
df_states['a2'] = ''

list1 = []
for i in range(len(df_states)):
    list1.append([[x, y] for x in range(df_states[['cumsum']].iloc[i][0]+2) for y in range(df_states[['cumsum']].iloc[i][0]+2)])

x = []
y = []
for i in range(len(list1)):
    for j in range(len(list1[i])):
        print(list1[i][j][0])
        x.append(list1[i][j][0])
        y.append(list1[i][j][1])

for i in range(len(df_states)):
    df_states['a1'].iloc[i] = x[i]
    df_states['a2'].iloc[i] = y[i]

print(df_states)

当然这不是最好的解决方案，但它对我有用

希望它也适合你。

祝你有个愉快的一天。

【讨论】：

实际上我需要 x1 x2 x3 cumsum a1 a2 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 1 而不是 x1 x2 x3 cumsum a1 a2 0 0 0 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 0 0 1 1 1

【解决方案2】：

我通过稍微调整 AMIR 的解决方案解决了我的问题。非常感谢阿米尔。

import pandas as pd

def multichoose(n,k):
    if k < 0 or n < 0: return "Error"
    if not k: return [[0]*n]
    if not n: return []
    if n == 1: return [[k]]
    return [[0]+val for val in multichoose(n-1,k)] + \
        [[val[0]+1]+val[1:] for val in multichoose(n,k-1)]

states=[]
for i in range(0,3):
    states=states+multichoose(3,i)
df_states = pd.DataFrame(states,columns=['x1','x2','x3'])
df_states = df_states.rename_axis('state_num').reset_index()
df_states['cumsum']=df_states['x1']+df_states['x2']+df_states['x3']
df_states

list1 = []
for i in range(len(df_states)):
    list1.append([[x, y] for x in range(df_states[['cumsum']].iloc[i][0]+2) for y in range(df_states[['cumsum']].iloc[i][0]+2)])

df_states['FIFO_LIFO']=list1
df_states.explode('FIFO_LIFO')

    state_num   x1  x2  x3  cumsum  FIFO_LIFO
0   0           0   0   0   0       [0, 0]
0   0           0   0   0   0       [0, 1]
0   0           0   0   0   0       [1, 0]
0   0           0   0   0   0       [1, 1]
1   1           0   0   1   1       [0, 0]
... ...         ... ... ... ...     ...
9   9           2   0   0   2       [2, 3]
9   9           2   0   0   2       [3, 0]
9   9           2   0   0   2       [3, 1]
9   9           2   0   0   2       [3, 2]
9   9           2   0   0   2       [3, 3]

【讨论】：