根据指定的属性分组成组

Question

我需要以这样的方式对数据进行分组，如果A1列的相邻值之间的差等于相同的预指定值，则它们属于同一组。如果两个相邻元素之间的值不同，则所有后续数据都属于另一组。例如，

import pandas as pd
import numpy as np

data = [
    [5, 2],
    [100, 23],
    [101, -2],
    [303, 9],
    [304, 4],
    [709, 14],
    [710, 3],
    [711, 3],
    [988, 21]
]
columns = ['a1', 'a2']
df = pd.DataFrame(data=data, columns=columns)

如果列A1的元素等于一个，则我有这样的数据表，则它们属于同一组，并且在此示例中的答案将如下：

[[0], [1, 2], [3, 4], [5, 6, 7], [8]]

输出列表存储与行相对应的索引来自DF。

排序A1列也可能有用。感谢您的帮助！

Answer 1

假设您的数据框是由a1排序的，并且我正确理解了您的问题，我认为您可以做这样的事情：

import pandas as pd
import numpy as np
from numba import njit

data = [
    [5, 2],
    [100, 23],
    [101, -2],
    [303, 9],
    [304, 4],
    [709, 14],
    [710, 3],
    [711, 3],
    [988, 21]
]
columns = ['a1', 'a2']
df = pd.DataFrame(data=data, columns=columns)

@njit
def get_groups(vals):
    counter = 0
    group = []
    for i in range(len(vals)-1):
        if vals[i+1]-vals[i] == 1:
            group.append(counter)
        else:
            group.append(counter)
            counter += 1
    if vals[-1] - vals[-2] == 1: group.append(group[-1])
    else: group.append(counter + 1)
        
    return group  
    
groups = get_groups(df['a1'].values)
assert len(groups) == len(df)

df['group'] = groups
final_ls = df.reset_index().groupby(['group']).agg({'index': list})['index'].to_list()
final_ls

------------------------------------------------------------
[[0], [1, 2], [3, 4], [5, 6, 7], [8]]
------------------------------------------------------------

njit decorator numba 使循环方法有效。

Answer 2

我们通过“ A1”列对数据框进行排序，然后找到相邻值的差异。现在我们有所不同，我们可以开始分组。

import pandas as pd

data = [
    [5, 2],
    [100, 23],
    [101, -2],
    [303, 9],
    [304, 4],
    [709, 14],
    [710, 3],
    [711, 3],
    [988, 21]
]
columns = ['a1', 'a2']
df = pd.DataFrame(data=data, columns=columns)

# To sort the values of "a1" column
df=df.sort_values(by=['a1'])
# To find the difference between the adjacent values
df['difference']=df['a1'].diff()

# Sorted index
indexs=df.index.tolist()

group=[]

# To check the difference, before this row
check=-1
for i,diff in enumerate(df['difference']):
    if diff==0 or diff==1:
        if check==1:
            group[-1].append(indexs[i])
        else:
            group.append([indexs[i-1],indexs[i]])
    check=diff

# For finding indexes that are not in group
z=[]
for x in group: [z.append(w) for w in x]
for t in (set(indexs)-set(z)):
    group.append([t])
print(group)

结果：

[[1, 2], [3, 4], [5, 6, 7], [0], [8]]

Answer 3

上面的答案促使我找到了一个简短而简单的代码以获取答案。非常感谢！

import pandas as pd
import numpy as np

data = [
    [5, 2],
    [100, 23],
    [101, -2],
    [303, 9],
    [304, 4],
    [709, 14],
    [710, 3],
    [711, 3],
    [988, 21]
]
columns = ['a1', 'a2']
df = pd.DataFrame(data=data, columns=columns)


diff = 1
index_groups = [[df.index[0]]]
for index in df.index[1:]:
    if df.loc[index, 'a1'] - df.loc[index-1, 'a1'] == diff:
        index_groups[-1].append(index)
    else:
        index_groups.append([index])
index_groups