将 dataframe 转换为元组列表

Question

我有一张桌子pandas DF看起来像

	奴隶	开始地址0	end_addr0	开始地址1	end_addr1	start_addr2	end_addr2
0	0	10000000	1FFFFFF	钠	钠	钠	钠
1	1	20000000	2007FFFF	40000000	40005FFF	钠	钠
2	1	20000000	2007FFFF	2010万	201FFFFF	钠	钠
3	2	20200000	202FFFF	20080000	20085FFF	40006000	400FFFF
4	3	0	0FFFFFF	钠	钠	钠	钠
5	4	20300000	203FFFF	钠	钠	钠	钠
6	5	2040万	204FFFFF	钠	钠	钠	钠

对于每个从属编号，我需要将其转换为范围列表（元组）。 例如，

Slave1_list = ( (20000000, 2007FFFF), (40000000, 40005FFF), (20100000, 201FFFFF))

从站（行）和地址对（列）的数量可以变化。

谢谢

编辑：

运行以下代码将样本数据加载到 dataframe 中：

import pandas as pd
import io

f = io.StringIO('''Slave|start_addr0|end_addr0|start_addr1|end_addr1|start_addr2|end_addr2
0|10000000|1FFFFFFF|NaN|NaN|NaN|NaN
1|20000000|2007FFFF|40000000|40005FFF|NaN|NaN
1|20000000|2007FFFF|20100000|201FFFFF|NaN|NaN
2|20200000|202FFFFF|20080000|20085FFF|40006000|400FFFFF
3|0|0FFFFFFF|NaN|NaN|NaN|NaN
4|20300000|203FFFFF|NaN|NaN|NaN|NaN
5|20400000|204FFFFF|NaN|NaN|NaN|NaN
''')
df = pd.read_csv(f, sep='|', engine='python', index_col=None)

Answer 1

如下所示：

import pandas as pd
from collections import defaultdict

data = [{'Slave': 1, 'start_addr0': 12, 'end_addr0': 189, 'start_addr1': 9, 'end_addr1': 17},
        {'Slave': 1, 'start_addr0': 3, 'end_addr0': 6, 'start_addr1': 1, 'end_addr1': 4},
        {'Slave': 3, 'start_addr0': 1, 'end_addr0': 7, 'start_addr1': 2, 'end_addr1': 14}]

df = pd.DataFrame(data)

print(df)
result = defaultdict(list)
rows = df.to_dict(orient='records')
for row in rows:
    slave = row.get('Slave')
    for key, start_value in row.items():
        if key.startswith('start_addr'):
            idx = key[-1]
            end_value = row.get('end_addr' + idx)
            result[slave].append((start_value, end_value))
        else:
            continue

print('result:')
print(result)

output

   Slave  start_addr0  end_addr0  start_addr1  end_addr1
0      1           12        189            9         17
1      1            3          6            1          4
2      3            1          7            2         14
result:
defaultdict(<class 'list'>, {1: [(12, 189), (9, 17), (3, 6), (1, 4)], 3: [(1, 7), (2, 14)]})

Answer 2

你可以试试：

通过wide_to_long一种选择：

df = df.reset_index()
result = pd.wide_to_long(df, stubnames=['start_addr', 'end_addr'], i=['index', 'Slave'], j='add_num', sep='').dropna(
).reset_index([0, -1], drop=True).apply(tuple, 1).groupby(level=0).agg(list)

通过groupby的一个选项：

k = df.set_index('Slave').stack().reset_index()
result = k.groupby(k.index//2).agg({'Slave': 'first', 0 : tuple}).groupby('Slave').agg({0 : set})

说明：

df.set_index('Slave').stack().reset_index()将删除NaN值并堆叠 dataframe。

k.groupby(k.index//2)将对备用行进行分组并执行所需的聚合（在此步骤中形成元组）

.groupby('Slave').agg({0: set}) -> 最后一个 groupby 是为每个从属捕获唯一的元组。

OUTPUT：

                                                                            0
Slave                                                                        
0                                                      {(10000000, 1FFFFFFF)}
1      {(40000000.0, 40005FFF), (20100000.0, 201FFFFF), (20000000, 2007FFFF)}
2      {(20080000.0, 20085FFF), (40006000.0, 400FFFFF), (20200000, 202FFFFF)}
3                                                             {(0, 0FFFFFFF)}
4                                                      {(20300000, 203FFFFF)}
5                                                      {(20400000, 204FFFFF)}

注意：我假设每个start_addr都存在一个end_addr 。

Answer 3

我认为这就是你要找的：

def make_tuples(x):
    return tuple([x['start_addr0'], x['end_addr0']])

# simple tuples
result = tuple(df[['start_addr0', 'end_addr0']].apply(make_tuples, axis=1).tolist())
print(result)

# unique tuples
unique_result = tuple(df[['start_addr0', 'end_addr0']].apply(make_tuples, axis=1).unique().tolist())
print(unique_result)

---

Output

((10000000, '1FFFFFFF'), (20000000, '2007FFFF'), (20000000, '2007FFFF'), (20200000, '202FFFFF'), (0, '0FFFFFFF'), (20300000, '203FFFFF'), (20400000, '204FFFFF'))
((10000000, '1FFFFFFF'), (20000000, '2007FFFF'), (20200000, '202FFFFF'), (0, '0FFFFFFF'), (20300000, '203FFFFF'), (20400000, '204FFFFF'))