英文:
Create pandas dataframe column with random conditional numbers
问题
我已创建了以下的pandas数据框。
import pandas as pd
import numpy as np
ds = {'col1' : [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2],
'col2' : [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]}
data = pd.DataFrame(data=ds)
它看起来像这样：
print(data)
col1 col2
0 1 0
1 1 0
2 1 0
3 1 0
4 1 0
5 1 0
6 1 0
7 1 0
8 1 0
9 1 0
10 1 0
11 1 0
12 1 0
13 1 0
14 2 1
15 2 1
16 2 1
17 2 1
18 2 1
19 2 1
20 2 1
21 2 1
22 2 1
23 2 1
24 2 1
25 2 1
26 2 1
27 2 1
我需要创建一个新的列（称为 col3），根据以下条件：
-
当
col1 = 1时，有14条记录，其中col2 = 0。新列（即col3）需要有14条记录中的50%的值等于col2（在这14条记录中随机分布），剩余的50%等于1。 -
当
col1 = 2时，有14条记录，其中col2 = 1。新列（即col3）需要有14条记录中的50%的值等于col2（在这14条记录中随机分布），剩余的50%等于0。
因此，最终的数据集将如下所示（请注意，col3 中的值的位置或记录是随机分配的）。
你需要的Python代码如下：
import random
# Define a function to generate col3 values based on the conditions
def generate_col3(row):
if row['col1'] == 1:
if row['col2'] == 0:
return [random.choice([0, 1]) for _ in range(14)]
else:
return [1] * 14
elif row['col1'] == 2:
if row['col2'] == 1:
return [random.choice([0, 1]) for _ in range(14)]
else:
return [0] * 14
# Apply the function to create the col3 column
data['col3'] = data.apply(generate_col3, axis=1)
# Explode the lists in col3 to separate rows
data = data.explode('col3', ignore_index=True)
# Shuffle the rows to randomize the order
data = data.sample(frac=1, random_state=42).reset_index(drop=True)
print(data)
这将生成符合您要求的数据框。
英文:
I have created the following pandas dataframe.
import pandas as pd
import numpy as np
ds = {'col1' : [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2],
'col2' : [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]}
data = pd.DataFrame(data=ds)
which looks like this:
print(data)
col1 col2
0 1 0
1 1 0
2 1 0
3 1 0
4 1 0
5 1 0
6 1 0
7 1 0
8 1 0
9 1 0
10 1 0
11 1 0
12 1 0
13 1 0
14 2 1
15 2 1
16 2 1
17 2 1
18 2 1
19 2 1
20 2 1
21 2 1
22 2 1
23 2 1
24 2 1
25 2 1
26 2 1
27 2 1
I need to create a new column (called col3) subject to the following conditions:
-
when
col1 = 1, there are 14 records for which col2 = 0. The new column (i.e.col3), needs to have 50% (of exactly those 14 records) of the values equal tocol2(randomly distributed across the 14 records) and the remaining 50% equal to 1. -
when
col1 = 2, there are 14 records for whichcol2 = 1. The new column (i.e.col3), needs to have 50% (of exactly those 14 records) of the values equal tocol2(randomly distributed across the 14 records) and the remaining 50% equal to 0.
So, the resulting dataset would look like this (bear in mind that the location - or record - of the values in col3 is randomly assigned):
Does anyone know the python code to produce such dataframe?
答案1
得分: 2
# 从col1的每个唯一值中随机抽取50%的样本到col3中
data['col3'] = data.groupby('col1')['col2'].sample(frac=.5)
# 使用预定义的col1值的映射填充剩余的50%
data['col3'] = data['col3'].fillna(data['col1'].map({1: 1, 2: 0}), downcast='infer')
英文:
groupby + sample
# take a sample of 50% from col2 per unique value in col1
data['col3'] = data.groupby('col1')['col2'].sample(frac=.5)
# fill the remaining 50% using a predefined mapping of col1 value
data['col3'] = data['col3'].fillna(data['col1'].map({1: 1, 2: 0}), downcast='infer')
Result
col1 col2 col3
0 1 0 1
1 1 0 0
2 1 0 0
3 1 0 0
4 1 0 0
5 1 0 0
6 1 0 0
7 1 0 1
8 1 0 0
9 1 0 1
10 1 0 1
11 1 0 1
12 1 0 1
13 1 0 1
14 2 1 1
15 2 1 0
16 2 1 1
17 2 1 0
18 2 1 0
19 2 1 1
20 2 1 1
21 2 1 0
22 2 1 1
23 2 1 0
24 2 1 0
25 2 1 1
26 2 1 1
27 2 1 0
答案2
得分: 1
我将为您翻译代码部分，以下是翻译好的内容：
# 使用df.sample()方法在条件内随机分离所有子组，然后分配值
ds = {'col1': [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2],
'col2': [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]}
data = pd.DataFrame(data=ds)
data['col_3'] = 0
# 设置数据帧并创建一个新的空列
# 现在让我们根据条件进行选择
cond_1 = data.loc[(data.col1 == 1) & (data.col2 == 0)]
cond_2 = data.loc[(data.col1 == 2) & (data.col2 == 1)]
# 获取第一个条件的随机50%，然后获取剩下的50%
cond_1_A = cond_1.sample(frac=0.5)
cond_1_B = cond_1.loc[cond_1.index.difference(cond_1_A.index)]
# 对于每个子组，将值设置为0或1
data.col_3.loc[cond_1_A.index] = 0
data.col_3.loc[cond_1_B.index] = 1
# 第二个条件 - 同样的操作
cond_2_A = cond_2.sample(frac=0.5)
cond_2_B = cond_2.loc[cond_2.index.difference(cond_2_A.index)]
data.col_3.loc[cond_2_A.index] = 0
data.col_3.loc[cond_2_B.index] = 1
# 完成
运行1：
col1 col2 col_3
0 1 0 0
1 1 0 1
2 1 0 1
3 1 0 0
4 1 0 1
5 1 0 1
6 1 0 0
7 1 0 0
8 1 0 0
9 1 0 1
10 1 0 0
11 1 0 0
12 1 0 1
13 1 0 1
14 2 1 1
15 2 1 1
16 2 1 0
17 2 1 0
18 2 1 0
19 2 1 0
20 2 1 1
21 2 1 0
22 2 1 0
23 2 1 1
24 2 1 1
25 2 1 1
26 2 1 0
27 2 1 1
运行2：
col1 col2 col_3
0 1 0 1
1 1 0 0
2 1 0 1
3 1 0 1
4 1 0 1
5 1 0 0
6 1 0 1
7 1 0 0
8 1 0 0
9 1 0 0
10 1 0 1
11 1 0 0
12 1 0 0
13 1 0 1
14 2 1 1
15 2 1 0
16 2 1 0
17 2 1 1
18 2 1 1
19 2 1 1
20 2 1 1
21 2 1 0
22 2 1 1
23 2 1 0
24 2 1 0
25 2 1 1
26 2 1 0
27 2 1 0
英文:
I would use the df.sample() method to isolate all the subgroups randomly within the condition, and then assign the values
ds = {'col1' : [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2],
'col2' : [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]}
data = pd.DataFrame(data=ds)
data['col_3'] = 0
sets up the dataframe and creates a new, empty column
Now let's select by condition
cond_1 = data.loc[(data.col1==1)&(data.col2==0)]
cond_2 = data.loc[(data.col1==2)&(data.col2==1)]
Get a random 50 % of the first condition, and then get the remaining 50%
cond_1_A = cond_1.sample(frac=.5)
cond_1_B = cond_1.loc[cond_1.index.difference(cond_1_A.index)]
For each sub-group, set the value to 0 or 1
data.col_3.loc[cond_1_A.index] = 0
data.col_3.loc[cond_1_B.index] = 1
Second Condition - Same Thing
cond_2_A = cond_2.sample(frac=.5)
cond_2_B = cond_2.loc[cond_2.index.difference(cond_2_A.index)]
data.col_3.loc[cond_2_A.index] = 0
data.col_3.loc[cond_2_B.index] = 1
That should do it.
Run 1
data
col1 col2 col_3
0 1 0 0
1 1 0 1
2 1 0 1
3 1 0 0
4 1 0 1
5 1 0 1
6 1 0 0
7 1 0 0
8 1 0 0
9 1 0 1
10 1 0 0
11 1 0 0
12 1 0 1
13 1 0 1
14 2 1 1
15 2 1 1
16 2 1 0
17 2 1 0
18 2 1 0
19 2 1 0
20 2 1 1
21 2 1 0
22 2 1 0
23 2 1 1
24 2 1 1
25 2 1 1
26 2 1 0
27 2 1 1
Run 2
data
col1 col2 col_3
0 1 0 1
1 1 0 0
2 1 0 1
3 1 0 1
4 1 0 1
5 1 0 0
6 1 0 1
7 1 0 0
8 1 0 0
9 1 0 0
10 1 0 1
11 1 0 0
12 1 0 0
13 1 0 1
14 2 1 1
15 2 1 0
16 2 1 0
17 2 1 1
18 2 1 1
19 2 1 1
20 2 1 1
21 2 1 0
22 2 1 1
23 2 1 0
24 2 1 0
25 2 1 1
26 2 1 0
27 2 1 0
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