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Python / Pandas - 基于几个变量和if / elif / else函数创建新变量

[英]Python/Pandas - creating new variable based on several variables and if/elif/else function

I am trying to create a new variable that is conditional based on values from several other values. 我正在尝试根据来自其他几个值的值创建一个有条件的新变量。 I'm writing here because I've tried writing this as a nested ifelse() statement in R, but it had too many nested ifelse's so it threw an error, and I think there should be an easier way to sort this out in Python. 我在这里写,因为我已经尝试将它写为R中的嵌套ifelse()语句,但它有太多嵌套的ifelse,所以它抛出了一个错误,我认为应该有一种更简单的方法来解决这个问题在Python中。

I have a dataframe (called df) that looks roughly like this (although in reality it's much bigger with many more month/year variables) that I've read in as a pandas DataFrame: 我有一个数据框(称为df)看起来大致像这样(虽然实际上它有更大的月份/年变量)我已经读作大熊猫DataFrame:

   ID  Sept_2015  Oct_2015  Nov_2015  Dec_2015  Jan_2016  Feb_2016  Mar_2016  \
0   1          0         0         0         0         1         1         1   
1   2          0         0         0         0         0         0         0   
2   3          0         0         0         0         1         1         1   
3   4          0         0         0         0         0         0         0   
4   5          1         1         1         1         1         1         1   

   grad_time  
0        240  
1        218  
2        236  
3          0  
4        206 

I'm trying to create a new variable that depends on values from all these variables, but values from "earlier" variables need to have precedent, so the if/elif/else condition would like something like this: 我正在尝试创建一个取决于所有这些变量的值的新变量,但“早期”变量的值需要有先例,所以if / elif / else条件需要这样的东西:

if df['Sept_2015'] > 0 & df['grad_time'] <= 236:
    return 236
elif df['Oct_2015'] > 0 & df['grad_time'] <= 237:
    return 237
elif df['Nov_2015'] > 0 & df['grad_time'] <= 238:
    return 238
elif df['Dec_2015'] > 0 & df['grad_time'] <= 239:
    return 239
elif df['Jan_2016'] > 0 & df['grad_time'] <= 240:
    return 240
elif df['Feb_2016'] > 0 & df['grad_time'] <= 241:
    return 241
elif df['Mar_2016'] > 0 & df['grad_time'] <= 242:
    return 242
else:
    return 0

And based on this, I'd like it to return a new variable that looks like this: 基于此,我希望它返回一个如下所示的新变量:

   trisk
0    240
1      0
2    240
3      0
4    236

I've tried writing a function like this: 我试过写这样的函数:

def test_func(df):
    """ Test Function for generating new value"""
    if df['Sept_2015'] > 0 & df['grad_time'] <= 236:
        return 236
    elif df['Oct_2015'] > 0 & df['grad_time'] <= 237:
        return 237
    ...
    else:
        return 0

and mapping it to the dataframe to create new variable like this: 并将其映射到数据框以创建新变量,如下所示:

new_df = pd.DataFrame(map(test_func, df)) 

However, when I run it, I get the following TypeError 但是,当我运行它时,我得到以下TypeError

 Traceback (most recent call last):

  File "<ipython-input-83-19b45bcda45a>", line 1, in <module>
     new_df = pd.DataFrame(map(new_func, test_df))

  File "<ipython-input-82-a2eb6f9d7a3a>", line 3, in new_func
     if df['Sept_2015'] > 0 & df['grad_time'] <= 236:

TypeError: string indices must be integers, not str

So I can see it's not wanting the column name here. 所以我可以看到它不希望列名在这里。 But I've tried this a number of other ways and can't get it to work. 但我已经尝试了许多其他方法,但无法让它发挥作用。 Also, I understand this might not be the best way to write this (mapping the function) so I am open to new ways to attempt to solve the problem of generating the trisk variable. 另外,我理解这可能不是写这个(映射函数)的最好方法,所以我愿意尝试解决生成trisk变量问题的新方法。 Thanks in advance and apologies if I haven't provided something. 如果我没有提供任何东西,请提前致谢并道歉。

Setup 建立

df = pd.DataFrame([[0, 0, 0, 0, 1, 1, 1, 240],
                   [0, 0, 0, 0, 0, 0, 0, 218],
                   [0, 0, 0, 0, 1, 1, 1, 236],
                   [0, 0, 0, 0, 0, 0, 0,   0],
                   [1, 1, 1, 1, 1, 1, 1, 206]],
                  pd.Index(range(1, 6), name='ID'),
                  ['Sept_2015', 'Oct_2015', 'Nov_2015', 'Dec_2015',
                   'Jan_2016', 'Feb_2016', 'Mar_2016', 'grad_time'])

I used mostly numpy for this 为此,我大多使用了numpy

a = np.array([236, 237, 238, 239, 240, 241, 242])
b = df.values[:, :-1]
g = df.values[:, -1][:, None] <= a

a[(b & g).argmax(1)] * (b & g).any(1)

Assigning it to new column 将其分配给新列

df['trisk'] = a[(b != 0).argmax(1)] * (b != 0).any(1)

df

在此输入图像描述

Without getting into streamlining your logic (which @piRSquared gets into): you can apply your test_func to the rows by issuing .apply(test_func, axis=1) to your dataframe. 无需简化逻辑(@piRSquared进入):您可以通过向数据帧发出.apply(test_func, axis=1)来将test_func应用于行。

import io
import pandas as pd

data = io.StringIO('''\
   ID  Sept_2015  Oct_2015  Nov_2015  Dec_2015  Jan_2016  Feb_2016  Mar_2016  grad_time  
0   1          0         0         0         0         1         1         1        240
1   2          0         0         0         0         0         0         0        218   
2   3          0         0         0         0         1         1         1        236
3   4          0         0         0         0         0         0         0          0
4   5          1         1         1         1         1         1         1        206
''')
df = pd.read_csv(data, delim_whitespace=True)

def test_func(df):
    """ Test Function for generating new value"""
    if df['Sept_2015'] > 0 & df['grad_time'] <= 236:
        return 236
    elif df['Oct_2015'] > 0 & df['grad_time'] <= 237:
        return 237
    elif df['Nov_2015'] > 0 & df['grad_time'] <= 238:
        return 238
    elif df['Dec_2015'] > 0 & df['grad_time'] <= 239:
        return 239
    elif df['Jan_2016'] > 0 & df['grad_time'] <= 240:
        return 240
    elif df['Feb_2016'] > 0 & df['grad_time'] <= 241:
        return 241
    elif df['Mar_2016'] > 0 & df['grad_time'] <= 242:
        return 242
    else:
        return 0

trisk = df.apply(test_func, axis=1)
trick.name = 'trisk'
print(trisk)

Output: 输出:

0    240
1      0
2    240
3      0
4    236
Name: trisk, dtype: int64

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