的Python：列表賦值超出範圍

Question

這個模塊是一個簡單的待辦事項應用程序，我與Python做的一部分......

def deleteitem(): 
      showlist() 
      get_item = int(raw_input("\n Enter number of item to delete: \n")) 
      f = open('todo.txt') 
      lines = f.readlines() 
      f.close() 
      lines[get_item] = "" 
      f = open('todo.txt','w') 
      f.writelines(lines) 
      f.close() 
      showlist() 

F中的行數明顯變化項目添加到列表中..這裏的問題是，例如如果在只有9個文件中的行（或其他任何不在範圍內），用戶輸入「10」，退出預期有：

IndexError: list assignment index out of range 

我可以添加到什麼該模塊，以便它提示用戶輸入範圍內的項目？我假設可能是一個嘗試塊...或者是有一種方法來捕捉異常..我猜測有一個簡單的方法來做到這一點...

Answer 1

明智更改當前代碼：

def deleteitem(): 
    showlist() 

    with open("todo.txt") as f: 
    lines = f.readlines() 
    if len(lines) == 0: # completely empty file 
    return # handle appropriately 
    prompt = "Enter number to delete (1-%d), or 0 to abort: " % len(lines) 
    while True: 
    input = raw_input(prompt) 
    try: 
     input = int(input, 10) 
    except ValueError: 
     print "Invalid input." 
    else: 
     if 0 <= input <= len(lines): 
     break 
     print "Input out of range." 
    if input == 0: 
    return 
    input -= 1 # adjust from [1,len] to [0,len) 

    #del lines[input] # if you want to remove that line completely 
    lines[input] = "\n" # or just make that line blank (what you had) 

    with open("todo.txt", "w") as f: 
    f.writelines(lines) 

    showlist() 

Answer 2

要麼捕捉IndexError索引或檢查len()的預先列出。

Answer 3

首先讀取該文件，然後要求用戶在一個循環，直到答案是可以接受的：

while True: 
    get_item = int(raw_input("\n Enter number of item to delete: \n")) 
    if get_item >=0 and get_item < len(lines): 
     break 

這將，當然，休息的時候該文件是空的，並沒有給出任何的暗示可接受的值給用戶。但讓我們爲你保留一些鍛鍊。

Answer 4

def deleteitem(): 
      showlist() 
      get_item = int(raw_input("\n Enter number of item to delete: \n")) 
      f = open('todo.txt') 
      lines = f.readlines() 
      f.close() 
      try: 
       lines[get_item] = "" 
      except Exception,err: 
       print err 
      f = open('todo.txt','w') 
      f.writelines(lines) 
      f.close() 
      showlist() 

Answer 5

嘗試是這樣的：

def deleteitem(): 

showlist() 
f = open('todo.txt') 
lines = f.readlines() 
f.close() 
if len(lines) == 0: 
    print "File is empty!" 
    return False 
print "File has %d items\n" % len(lines) 
item = 0 
while item < len(lines): 
    item = raw_input("\n Enter number of item to delete(0-%d): \n" % len(lines)) 
    item = int(item) # because of the width of the code 
f = open('todo.txt','w') 
f.write(lines[0:item-1]) 
f.write(lines[item::]) 
f.close() 
showlist() 

Answer 6

對於它的價值....我把代碼給我的待辦事項.py程序在這裏......這只是我從OS X的一個終端運行來保持我在工作中需要做的事情的控制......我確定它是可怕的非pythonic，低效率和其他一切......但也許這對某個在這個線程中發現的人會有用處：

from __future__ import with_statement 
import sys 
import os 
import fileinput 

os.system('clear') 

print ("##############   TO DO LIST  ############") 
print ("##############       ############") 

def showlist(): 
    os.system('clear') 
    print ("############ Current To Do List ######") 
    print ("########################################") 

    get_list = open('todo.txt') 
    entire_list = get_list.readlines() 
    for i in range (len(entire_list)): 
     print i, entire_list[i] 
    get_list.close() 
    print ("########################################") 
    print ("########################################") 

def appendlist(): 
    print ("#######################################") 
    print ("#######################################") 


    addtolist = str(raw_input("Enter new item: \n")) 
    thelist = open('todo.txt', 'a') 
    thelist.write(str(addtolist)) 
    thelist.write(str('\n')) 
    thelist.close() 
    showlist() 


def deleteitem(): 
    showlist() 

     with open("todo.txt") as f: 
      lines = f.readlines() 
      if len(lines) == 0: 
       return 
     prompt = "Enter number to delete or '0' to abort: " 
     while True: 
       input = raw_input(prompt) 
       try: 
        input = int(input, 10) 
       except ValueError: 
        print "Invalid input." 
       else: 
        if 0 <= input <= len(lines): 
         break 
        print "Input out of range." 
     if input == 0: 
        return 

     lines[input] = "" 

      with open("todo.txt", "w") as f: 
       f.writelines(lines) 

     showlist() 

while True: 

    askme = raw_input("\nDo you want to:\n(S)ee list\n(A)ppend list\n(D)elte from list\n(Q)Quit?\n") 
    print str('\n') 

    if askme == "S": 
     showlist() 
    elif askme == "A": 
     appendlist() 
    elif askme == "D": 
     deleteitem() 

    elif askme == "Q": 
     sys.exit() 
    else: 
     print ("Try again?") 

print ("#######################################") 
print ("#######################################")