将数据框转换为元组列表

Posted 2023-02-23

【中文标题】将数据框转换为元组列表

【英文标题】：Convert a dataframe to a list of tuples

【发布时间】：2021-08-11 22:05:24

【问题描述】：

我有一张桌子 pandas DF，看起来像

	Slave	start_addr0	end_addr0	start_addr1	end_addr1	start_addr2	end_addr2
0	0	10000000	1FFFFFFF	NaN	NaN	NaN	NaN
1	1	20000000	2007FFFF	40000000	40005FFF	NaN	NaN
2	1	20000000	2007FFFF	20100000	201FFFFF	NaN	NaN
3	2	20200000	202FFFFF	20080000	20085FFF	40006000	400FFFFF
4	3	0	0FFFFFFF	NaN	NaN	NaN	NaN
5	4	20300000	203FFFFF	NaN	NaN	NaN	NaN
6	5	20400000	204FFFFF	NaN	NaN	NaN	NaN

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))

从属设备（行）和地址对（列）的数量可以变化。

谢谢

编辑：

运行以下代码将示例数据加载到数据框中：

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)

【问题讨论】：

你能发一个我们可以copy and run的df吗。其次，给出该数据帧的确切预期输出。谢谢

【参考方案1】：

类似于以下内容：

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)

输出

   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)])

【参考方案2】：

我想这就是你要找的：

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)

---

输出

((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'))

【参考方案3】：

你可以试试：

一个选项通过wide_to_long:

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)

一个选项来自groupby：

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

解释：

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

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。

输出：

                                                                            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。