implementation and "ISSUE" colle

1. when copy file with shutil and os.listdir()

IsADirectoryError: [Errno 21]

• The problem comes from the use of os.listdir() returns both files and folders which causes the confusion for shutil.copy() function.
• The resolution is the use of os.walk() function instead of os.listdir()

2. How to copy folder from source to deset in python.

• The why to copy folder form one place to another place is shutil.copytree.
• The key usage of this function is to make sure the dest location is a completely new location.
• Demo code for copying one folder with many of the sub-folders into another location

import pandas as pd
import os
import shutil

file_path    = '../DSB3/stage1'
target_path  = '/media/loveplay1983/dl_lab/thesis/LungNoduleDetectionClassification/DSB3/malignancies'
csv_path     = '../DSB3'

malig_tbl    =  pd.concat([pd.read_csv(csv_path+'/stage1_labels_1397.csv'), \
                         pd.read_csv(csv_path+'/stage1_solution_198.csv')])
malig_tbl.head()

patients     = malig_tbl['id'].values
status       = malig_tbl['cancer'].values

print(patients, status)
nods_file    = os.listdir(file_path)
nods_index = []

for i, each in enumerate(status):
  if each == 1:
      nods_index.append(i)
  
nodes = malig_tbl.iloc[nods_index]

nodes_true = []
for each in nods_file:
  for i, r in nodes.iterrows():
      if each == r.id:
          nodes_true.append(each)
          
print(len(nodes_true))
inx = 0
while inx != 419:
  for each in nodes_true:
      shutil.copytree(file_path+'/{}'.format(each), target_path+'/nods_{}'.format(inx))
      inx += 1  

2. Python create and delete file demo

#!/usr/bin/python
#-*-coding:utf-8-*-       #指定编码格式，python默认unicode编码
 
import os
directory = "./dir"
os.chdir(directory)  #切换到directory目录
cwd = os.getcwd()  #获取当前目录即dir目录下
print("------------------------current working directory------------------")
 
def deleteBySize(minSize):
    """删除小于minSize的文件（单位：K）"""
    files = os.listdir(os.getcwd())  #列出目录下的文件
    for file in files:
        if os.path.getsize(file) < minSize * 1000:
            os.remove(file)    #删除文件
            print(file + " deleted")
    return
 
def deleteNullFile():
    '''删除所有大小为0的文件'''
    files = os.listdir(os.getcwd())
    for file in files:
        if os.path.getsize(file)  == 0:   #获取文件大小
            os.remove(file)
            print(file + " deleted.")
    return
 
def create():
    '''根据本地时间创建新文件，如果已存在则不创建'''
    import time
    t = time.strftime('%Y-%m-%d',time.localtime())  #将指定格式的当前时间以字符串输出
    suffix = ".docx"
    newfile= t+suffix
    if not os.path.exists(newfile):
        f = open(newfile,'w')
        print newfile
        f.close()
        print newfile + " created."
    else:
        print newfile + " already existed."
    return
hint = '''funtion:
            1   create new file
            2   delete null file
            3   delete by size
please input number:'''
while True:
    option = raw_input(hint)  #获取IO输入的值
    if cmp(option,'1') == 0:
        create()
    elif cmp(option,'2') == 0:
        deleteNullFile()
    elif cmp(option,'3') == 0:
        minSize = raw_input("minSize(K):")
        deleteBySize(minSize)
    elif cmp(option,'q') == 0:
        print "quit !"
        break
    else:
        print ("disabled input ,please try again....")

3. Lamda expression (转自https://www.cnblogs.com/hf8051/p/8085424.html)

• 应用在函数式编程中

Python提供了很多函数式编程的特性，如：map、reduce、filter、sorted等这些函数都支持函数作为参数，lambda函数就可以应用在函数式编程中。如下：

需求：将列表中的元素按照绝对值大小进行升序排列

list1 = [3,5,-4,-1,0,-2,-6]
sorted(list1, key=lambda x: abs(x))
当然，也可以如下：

list1 = [3,5,-4,-1,0,-2,-6]
def get_abs(x):
    return abs(x)
sorted(list1,key=get_abs)

只不过这种方式的代码看起来不够Pythonic

• 应用在闭包中

def get_y(a,b):
     return lambda x:ax+b
y1 = get_y(1,1)
y1(1) # 结果为2

当然，也可以用常规函数实现闭包

def get_y(a,b):
    def func(x):
        return ax+b
    return func
y1 = get_y(1,1)
y1(1) # 结果为2

只不过这种方式显得有点啰嗦

那么是不是任何情况下lambda函数都要比常规函数更清晰明了呢？
肯定不是。
Python之禅中有这么一句话：Explicit is better than implicit（明了胜于晦涩），就是说那种方式更清晰就用哪一种方式，不要盲目的都使用lambda表达式。