[英]Convert a dataframe to a list of tuples
I have a table pandas DF which looks like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![]() |
For each slave number I need to convert it to a list of ranges (tuples).对于每个从属编号,我需要将其转换为范围列表(元组)。 For example,
例如,
Slave1_list = ( (20000000, 2007FFFF), (40000000, 40005FFF), (20100000, 201FFFFF))
The number of slaves (rows) and address-pairs (columns) can vary.从站(行)和地址对(列)的数量可以变化。
Thanks谢谢
EDIT :编辑:
Run the following code to load sample data into dataframe: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)
Something like the below: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 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)]})
You can try:你可以试试:
One option via wide_to_long
: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)
An option via groupby
: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})
Explanation :说明:
df.set_index('Slave').stack().reset_index()
will remove the NaN
values and stack the dataframe. df.set_index('Slave').stack().reset_index()
将删除NaN
值并堆叠 dataframe。
k.groupby(k.index//2)
will group alternate rows and perform the required aggregations(tuples are formed in this step) 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})
-> Last groupby is to capture the unique tuples for each slave. .groupby('Slave').agg({0: set})
-> 最后一个 groupby 是为每个从属捕获唯一的元组。
OUTPUT: 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)}
NOTE: I'm assuming for every start_addr
there exists an end_addr
.注意:我假设每个
start_addr
都存在一个end_addr
。
I think this is what you are looking for: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 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'))
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