如何将文件目录作为方法参数传递？

Question

这里的代码是我首先读取文件配置，然后我需要将目录传递到称为ednotes的方法中_提取器。获取注释（）

 from preprocessing import ednotes_extractor

    csf_config_txt = open("..\CFS_Config.txt", "r")

    file_list = []
    root_dir = ""
    edn_txt = ""
    pcr_txt = ""

    for line in csf_config_txt:
        file_list.append(line)

    if "Folder Path = " in file_list[0]:
        root_dir = str(file_list[0])
        root_dir = root_dir.replace("Folder Path = ", "")
        root_dir = root_dir.replace("\n", "")

    if "ED Notes = " in file_list[1]:
        root_dir = str(file_list[1])
        root_dir = root_dir.replace("ED Notes = ", "")
        root_dir = root_dir.replace("\n", "")

    if "Patient Care Record = " in file_list[2]:
        root_dir = str(file_list[2])
        root_dir = root_dir.replace("Patient Care Record = ", "")
        root_dir = root_dir.replace("\n", "")

    print(file_list[2])

    def convert_txt(choices):
        if(choices == 1):
            #I would like to pass a directory in this method called theednotes_extractor 
            ednotes_extractor.get_ednotes(str2=file_list[1])
        elif(choices==2):
            print("Hi")

Answer 1

也许这会有帮助

arr1 = [1,2,3]
arr2 = [4,5,6]

只需添加两个数组而不使用[]

arr3 = arr1 + arr2

对你来说

x = []
for i in range(0, len(arr_cat)):
    x += arr_cat[i]
print(len(x))

以上就足够了。你知道吗

Answer 2

试试这个：

from itertools import chain

new_list = list(chain(*arr_cat))

Answer 3

列表理解就可以了。它将打开列表列表并将其放入新列表中。你知道吗

flat_list = [item for sublist in arr_cat for item in sublist]
arr_cat = flat_list