在列列表上应用自定义函数

Question

我正在尝试通过应用自定义函数来优化一些工作代码，但我不确定如何对大型数据框中的特定列进行优化。在下面的示例中，我在我的数据框中选择开放式问题，这是一项调查。您会看到我手动输入每个开放式列，但我只想有一个迭代开放式列表的自定义函数。

openend = ['Q28','Q56','Q63']

### Change ranges to match the above
open1 = df.iloc[:, 28:29] # isolates 'range'
open1 = open1.iloc[1:] # removes first row
open1 = pd.concat([ids, open1], axis=1) # adds ids

open2 = df.iloc[:, 56:57]
open2 = open2.iloc[1:]
open2 = pd.concat([ids, open2], axis=1)

open3 = df.iloc[:, 63:64]
open3 = open3.iloc[1:]
open3 = pd.concat([ids, open3], axis=1)

open1['question'] = df1['Q28'][0]
open1['answer'] = open1.iloc[:,1:2]
open1 = open1.drop(open1.iloc[:,1:2], axis=1)

open2['question'] = df1['Q56'][0]
open2['answer'] = open2.iloc[:,1:2]
open2 = open2.drop(open2.iloc[:,1:2], axis=1)

open3['question'] = df1['Q63'][0]
open3['answer'] = open3.iloc[:,1:2]
open3 = open3.drop(open3.iloc[:,1:2], axis=1)

open1_stack = open1
open2_stack = open2
open3_stack = open3

open1_stack["answer"] = open1_stack["answer"].str.upper().str.title()
open1_count = open1_stack.answer.str.split(expand=True).stack().value_counts()
open1_count = open1_count.to_frame().reset_index()
open1_count.columns = ['Word', 'Count']
open1_count['question'] = df1['Q28'][0]

open2_stack["answer"] = open2_stack["answer"].str.upper().str.title()
open2_count = open2_stack.answer.str.split(expand=True).stack().value_counts()
open2_count = open2_count.to_frame().reset_index()
open2_count.columns = ['Word', 'Count']
open2_count['question'] = df1['Q56'][0]

open3_stack["answer"] = open3_stack["answer"].str.upper().str.title()
open3_count = open3_stack.answer.str.split(expand=True).stack().value_counts()
open3_count = open3_count.to_frame().reset_index()
open3_count.columns = ['Word', 'Count']
open3_count['question'] = df1['Q63'][0]

有人可以通过这个示例向我展示如何遍历开放式列表并以最佳方式应用这些函数吗？

提前致谢。

Answer 1

您可以使用接受openend 作为参数并具有如下签名的函数来包装所有代码：

def prepare_survey(openend:list):

然后遍历该列表以提取'QXX'：

for q in openend:
    # process

我看到您使用的内容与 openend 相同，除了提取索引的前几个步骤外没有任何变化。所以，保持原样，但提取问题编号，如下所示：

import re

def prepare_survey(openend:list):
    for q in openend:
        # process
        idx = int(re.sub("[^0-9]", "", q))  # extract question number
        # continue with the steps you have
        open1 = df.iloc[:, idx:idx+1]