I recently started to study Python, and as I was trying to run a code from a book (with my modification) I got the error:
IndexError: list assignment index out of range
in : `Names[len(Names)]=name`
I read some questions with this error on web but can't figure it out.
Names=[]
num=0
name=''
while True :
print('Enter the name of person '+str(len(Names)+1) + '(or Enter nothing to stop)')
name=input()
if name == '' :
break
Names[len(Names)]=name
print('the person names are:')
for num in range(len(Names)+1) :
print(' '+Names[num])
Looks like you want to append something to an existing list. Why not use .append()? This won't give you the IndexError.
Names.append(name)
Another same error: You shouldn't write range(len(Names) + 1). range(len(Names)) is enough for you to iterate through the whole list:
for num in range(len(Names)):
print(' '+Names[num])
Another suggestion: You don't need the for loop to print the result, at all. Just use str.join():
print(' '.join(Names))
You can not access out of range index
Ex:
>>> l = [1,2,3]
>>> l = [0,1,2]
>>> l[3] = "New"
Traceback (most recent call last):
File "<pyshell#2>", line 1, in <module>
l[3] = "New"
IndexError: list assignment index out of range
For that, you have to append new data to the list.
>>> l.append("new")
>>> l
[0, 1, 2, 'new']
You can try:
Names=[]
num=0
name=''
while True :
print('Enter the name of person '+str(len(Names)+1) + '(or Enter nothing to stop)')
name=input()
if name == '' :
break
Names.append(name)
print('the person names are:')
for num in range(len(Names)) :
print(' '+Names[num])
just use the append function to insert the new name to the existing list of names.
Syntax:-
if you want to append 'foo' to the list of existing names i.e 'Names',type Names.append('foo').
Related
Alien Dictionary
Link to the online judge -> LINK
Given a sorted dictionary of an alien language having N words and k starting alphabets of standard dictionary. Find the order of characters in the alien language.
Note: Many orders may be possible for a particular test case, thus you may return any valid order and output will be 1 if the order of string returned by the function is correct else 0 denoting incorrect string returned.
Example 1:
Input:
N = 5, K = 4
dict = {"baa","abcd","abca","cab","cad"}
Output:
1
Explanation:
Here order of characters is
'b', 'd', 'a', 'c' Note that words are sorted
and in the given language "baa" comes before
"abcd", therefore 'b' is before 'a' in output.
Similarly we can find other orders.
My working code:
from collections import defaultdict
class Solution:
def __init__(self):
self.vertList = defaultdict(list)
def addEdge(self,u,v):
self.vertList[u].append(v)
def topologicalSortDFS(self,givenV,visited,stack):
visited.add(givenV)
for nbr in self.vertList[givenV]:
if nbr not in visited:
self.topologicalSortDFS(nbr,visited,stack)
stack.append(givenV)
def findOrder(self,dict, N, K):
list1 = dict
for i in range(len(list1)-1):
word1 = list1[i]
word2 = list1[i+1]
rangej = min(len(word1),len(word2))
for j in range(rangej):
if word1[j] != word2[j]:
u = word1[j]
v = word2[j]
self.addEdge(u,v)
break
stack = []
visited = set()
vlist = [v for v in self.vertList]
for v in vlist:
if v not in visited:
self.topologicalSortDFS(v,visited,stack)
result = " ".join(stack[::-1])
return result
#{
# Driver Code Starts
#Initial Template for Python 3
class sort_by_order:
def __init__(self,s):
self.priority = {}
for i in range(len(s)):
self.priority[s[i]] = i
def transform(self,word):
new_word = ''
for c in word:
new_word += chr( ord('a') + self.priority[c] )
return new_word
def sort_this_list(self,lst):
lst.sort(key = self.transform)
if __name__ == '__main__':
t=int(input())
for _ in range(t):
line=input().strip().split()
n=int(line[0])
k=int(line[1])
alien_dict = [x for x in input().strip().split()]
duplicate_dict = alien_dict.copy()
ob=Solution()
order = ob.findOrder(alien_dict,n,k)
x = sort_by_order(order)
x.sort_this_list(duplicate_dict)
if duplicate_dict == alien_dict:
print(1)
else:
print(0)
My problem:
The code runs fine for the test cases that are given in the example but fails for ["baa", "abcd", "abca", "cab", "cad"]
It throws the following error for this input:
Runtime Error:
Runtime ErrorTraceback (most recent call last):
File "/home/e2beefe97937f518a410813879a35789.py", line 73, in <module>
x.sort_this_list(duplicate_dict)
File "/home/e2beefe97937f518a410813879a35789.py", line 58, in sort_this_list
lst.sort(key = self.transform)
File "/home/e2beefe97937f518a410813879a35789.py", line 54, in transform
new_word += chr( ord('a') + self.priority[c] )
KeyError: 'f'
Running in some other IDE:
If I explicitly give this input using some other IDE then the output I'm getting is b d a c
Interesting problem. Your idea is correct, it is a partially ordered set you can build a directed acyclcic graph and find an ordered list of vertices using topological sort.
The reason for your program to fail is because not all the letters that possibly some letters will not be added to your vertList.
Spoiler: adding the following line somewhere in your code solves the issue
vlist = [chr(ord('a') + v) for v in range(K)]
A simple failing example
Consider the input
2 4
baa abd
This will determine the following vertList
{"b": ["a"]}
The only constraint is that b must come before a in this alphabet. Your code returns the alphabet b a, since the letter d is not present you the driver code will produce an error when trying to check your solution. In my opinion it should simply output 0 in this situation.
I am going through the book Automate the Boring Stuff with Python, and need help understanding what's really happening with the code.
catNames = []
while True:
print('Enter the name of cat ' + str(len(catNames) + 1) + ' (Or enter nothing to stop.):')
name = input()
if name == '':
break
catNames = catNames + [name]
print('The cat names are:')
for name in catNames:
print(' ' + name)
now it makes sense until
for name in catNames:
print(' ' + name)
I am only used to seeing for loops with range(), and this does not make sense to me. A detailed explanation would be highly appreciated thanks
I will explain it to you on a simple example:
# You create some list of elements
list = [1, 2, 3, 9, 4]
# Using for loop you print out all the elements
for i in list:
print(i)
It will print to the console:
1
2
3
9
4
And you can also do it by using range but you have to know the length of the array:
# You create some list of elements
list = [1, 2, 3, 9, 4]
# get the list length
length = len(list)
# Iterating the index
# same as 'for i in range(len(list))'
for i in range(length):
print(list[i])
Console output will look the same as before
I am learning Python. I tried to get the keys of a dictionary. But I only get the last key. In my understanding, method keys() is used to get all keys in the dictionary.
Following are my questions?
1. Why I cannot get all keys?
2. If I have a dictionary, how can I get the value if I know the key? e.g. dict = {'Ben':8, 'Joe':7, 'Mary' : 9}. How can I input the key = "Ben", so the program can output the value 8? The tutorial shows that the key must be immutable. This constraint is very inconvenient when trying to get a value with a given key.
Any suggestion would be highly appreciated.
Here are my code.
import os, tarfile, urllib
work_path = os.getcwd()
input_control_file = "input_control"
import os, tarfile, urllib
work_path = os.getcwd()
input_control_file = "input_control"
input_control= work_path + "/" + input_control_file
#open control file if file exist
#read setting info
try:
#if the file does not exist,
#then it would throw an IOError
f = open(input_control, 'r')
#define dictionary/hash table
for LINE in f:
LINE = LINE.strip() #remove leading and trailing whitespace
lst = LINE.split() #split string into lists
lst[0] = lst[0].split(":")[0]
dic = {lst[0].strip():lst[1].strip()}
except IOError:
# print(os.error) will <class 'OSError'>
print("Reading file error. File " + input_control + " does not exist.")
#get keys
def getkeys(dict):
return list(dict.keys())
print("l39")
print(getkeys(dic))
print("end")
Below are the outputs.
l39
['source_type']
end
The reason is that you are reassigning variable dic again in for loop. You are not updating or adding the dictionary, instead you are reassigning the variable. In that case, dic will have only the last entry. You can change your for loop to:
dic = {}
for LINE in f:
LINE = LINE.strip() #remove leading and trailing whitespace
lst = LINE.split() #split string into lists
lst[0] = lst[0].split(":")[0]
dic.update({lst[0].strip():lst[1].strip()}) # update the dictionary with new values.
For your other question, if you have the dictionary dic = {'Ben':8, 'Joe':7, 'Mary' : 9}, then you can get the value by: dic['Ben']. It will return the value 8 or will raise KeyError if key Ben is not found in the dictionary. To avoid KeyError, you can use the get() method of dictionary. It will return None if provided key is not found in the dictionary.
val = dic['Ben'] # returns 8
val = dic['Hen'] # will raise KeyError
val = dic.get('Hen') # will return None
In your for loop, you are re-initializing the dictionary value, while you need to update the dictionary, i.e., append the key-value pair to the pre-existing dictionary. For this, use
dic.update({lst[0].strip() : lst[1].strip()})
This will update the key-value pair to the dictionary. Now, when you use dic.keys(), you will get all the keys of dic, as a list.
As for your second question, access the dictionary, just like accessing a list, except that list is accessed with indices, and dictionary will be accessed by keys. Say, you have a list and a dictionary as
lst = [1, 2, 3, 4, 5]
dic = {'a' : 1, 'b' : 2, 'c' : 3, 'd' : 4, 'e' : 5}
To get value 2 from list, you do lst[1], i.e., value at index 1. Similarly, if you want to get the value 2 from dictionary, do dic['b'], i.e., value of key 'b'. It is as simple as that.
I'm trying to make a program that will pick up randomly a name from a file. The user would be asked if he wants to pick up another one again (by pressing 1).
The names can't be picked up twice.
Once picked up, the names would be stocked in a list, written into a file.
When all the names are picked up, the program would be able to restart from the beginning.
I checked other similar problems, but I still don't get it...
from random import *
#import a list of name from a txt file
def getL1():
l1 = open("Employees.txt", "r")
list1 = []
x = 0
for line in l1:
list1.append(line)
x = x+1
list1 = [el.replace('\n', '') for el in list1]
#print("list" 1 :",list)
return list1
#import an empty list (that will be filled by tested employees) during
#execution of the program
def getL2():
l2 = open("tested.txt", "r")
list2 = []
for line in l2:
list2.append(line)
list2 = [el.replace('\n', '') for el in list2]
#print("list 2 :",list2)
l2.close()
return list2
def listCompare():
employees = getL1()#acquire first list from employee file
tested = getL2()#acquire second list from tested file
notTested = []# declare list to hole the results of the compare
for val in employees:
if val not in tested: #find employee values not present in tested
#print(val)
notTested.append(val)#append value to the notTested list
return notTested
def listCount():
x=0
employees = getL1()
tested = getL2()
for val in employees:
if val not in tested:
x = x+1
return x
#add the names of tested employees the the second second file
def addTested(x):
appendFile = open("tested.txt", "a")
appenFile.write(x)
appendFile.write('\n')
appendFile.close()
def main():
entry = 1
while entry == 1:
pickFrom = listCompare()
if listCount() > 0:
y = randint (0, (listCount ()-1))
print ('\n' + "Random Employee to be tested is: ", pickFrom(y), "\n")
addTested(pickFrom[y])
try:
entry = int(input("Would you like to test another employee? Enter 1:"))
except:
print("The entry must be a number")
entry = 0
else:
print("\n/\ new cycle has begun")
wipeFile = open("tested.txt", "w")
print ("goodbye")
main()
The last error that I have is :
Traceback (most recent call last):
File "prog.py", line 78, in <module>
main()
File "prog.py", line 65, in main
print ('\n' + "Random Employee to be tested is: ", pickFrom(y), "\n")
TypeError: 'list' object is not callable
As per the code print pickFrom is a list and when you are referencing it in the print it needs to be called using [ ]. Change it to pickFrom[y]
I'm working out how to remove a specific item from a list.
"peppers", "cheese", "mushrooms", "bananas", "peppers"
I can locate the item "peppers", and change it to "gone!", but I really want to deleting the item, using
del blist[idx]
But that causes an error and I don't know why.
myList = ["peppers", "cheese", "mushrooms", "bananas", "peppers"]
def findInList (needle, haystack):
needle = needle.lower()
findy = []
# loops over list elements
for i in range(0, len(haystack)):
temp = haystack[i].lower()
idx = temp.find(needle)
if (idx != -1): findy.append(i)
return findy
def deleteItemInList(alist, blist):
for i in range(0, len(alist)):
idx = alist[i]
blist[idx] = "Gone!"
# del blist[idx]
# find items in list
mySearch = findInList("Peppers", myList)
# remove item from list
deleteItemInList(mySearch, myList)
print myList
Traceback: as follows
Traceback (most recent call last):
File "delete_in_list.py", line 23, in <module>
deleteItemInList(mySearch, myList)
File "delete_in_list.py", line 16, in deleteItemInList
blist[idx] = "Gone!"
IndexError: list assignment index out of range
Could someone look over the code above and point out where I'm going wrong.
You can use a list comprehension for this.
def removeWord(needle, haystack):
return [word for word in haystack if word.lower() != needle.lower()]
To find an element use this function. Or alternatively just define it as usual:
>>> find = lambda _list, query: [item.lower() for item in _list].index(query.lower())
>>> l = ['red', 'pepper']
>>> q = 'Pepper'
>>> find(l, q)
1
To remove by index just use del:
>>> del l[find(l, q)]
>>> l
['red']
I finally figured it out! Whilst iterating over the list deleting items in the list I was effectively sawing off the branch I was sitting on.
You need to loop over the list in reverse:
def deleteItemInList(alist, blist):
for i in range(len(alist) -1, -1, -1):
idx = alist[i]
del blist[idx]