I've looked around at other posts on here for a solution to my problem but none of them seems to work for me. I want to append a tuple to (what starts out as) an empty list. I've checked my code and it looks like whenever I try to just append my tuple it turns the list into a NoneType object. Here's what I've got:
list_of_pairs = []
for player in standings:
opponent = standings[standings.index(player) + 1]
matchup = (player[0:2] + opponent[0:2])
list_of_pairs = list_of_pairs.append(matchup)
print "ADDED TO LIST: " + str(list_of_pairs)
Each player in the standings list contains a tuple with four elements. I've tried using index and re-assigning list_of_pairs = [list_of_pairs, matchup], but nothing seems to be returning the right thing (i.e. [(player),(next player), (another player), (etc.)]. Every time I print "added to list" I just get ADDED TO LIST: None. I've checked matchup as well and it's definitely storing the first two values of the respective players fine. Any ideas?
This is because appending an element to a list returns None, which you are then printing when you do:
list_of_pairs = list_of_pairs.append(matchup)
print "ADDED TO LIST: " + str(list_of_pairs)
For example:
>>> a = []
>>> b = a.append('hello')
>>> print a
['hello']
>>> print b
None
Related
Given the python 2.7 code below:
# faultReportObj = {}
flatDataObj = []
while row:
faultReportObj = {}
faultReportObj['MessageTime'] = row['MessageTime']
faultReportObj['Event'] = row['Event']
faultReportObj['Subsystem'] = row['Subsystem']
faultReportObj['UnifiedFaultCode'] = row['UnifiedFaultCode']
faultReportObj['FaultDescription'] = row['FaultDescription']
faultReportObj['FaultDetails'] = row['FaultDetails']
faultReportObj['FirstOccurrenceDateTime'] = row['FirstOccurrenceDateTime']
faultReportObj['LastOccurrenceDateTime'] = row['LastOccurrenceDateTime']
faultReportObj['OccurrenceCount'] = row['OccurrenceCount']
print "current row:"
pp.pprint(faultReportObj)
flatDataObj.append(faultReportObj)
row = cursor.fetchone()
conn.close()
pp.pprint(flatDataObj)
If I declare faultReportObj outside the while loop, I get (say) 96 entries in flatDataObj that are all identical to the very last row returned by the query. Note that the pprint statement within the while loop prints the expected (varying) results.
If, as above, I declare faultReportObj inside the loop, flatDataObj is loaded correctly.
Why???? Why is the very last row returned being propagated throughout the entire list?
Thanks!
This is due to list.append inserting a reference to faultReportObj and not copying the value of the dict.
Another way of looking at it:
If you define faultReportObj before the loop, the following occurs:
Create a new dict.
Populate it.
Append a reference to the dict into the list.
Change the dict's content.
Append another reference to the same dict.
etc.
Here's a short piece of code that exemplifies this property:
>>> d = {}
>>> l = []
>>> l.append(d)
>>> d[1] = 2
>>> l
[{1: 2}]
What you want is for step one (Create a new dict) to happen in every iteration of the loop, so that you append a different dict every time.
So I am rather new to programming and just recently started with Classes and we are supposed to make a phonebook that can be loaded in seperate text files.
I however keep running into the problem in this section that when I get into the for-loop. It hits a brick wall on
if storage[2] == permaStorage[i].number:
And tells me "IndexError: list index out of range". I am almost certain it is due to permaStorage starts out empty, but even when I attempt to fill it with temporary instances of Phonebook it tells me it out of range. The main reason it is there is to check if a phone number already exists within the permaStorage.
Anyone got a good tip on how to solve this or work around it?
(Sorry if the text is badly written. Just joined this site and not sure on the style)
class Phonebook():
def __init__(self):
self.name = ''
self.number = ''
def Add(name1, number1):
y = Phonebook()
y.name = name1
y.number = number1
return y
def Main():
permaStorage = []
while True:
print " add name number\n lookup name\n alias name newname\n change name number\n save filename\n load filename\n quit\n"
choices = raw_input ("What would you like to do?: ")
storage = choices.split(" ")
if storage[0] == "add":
for i in range(0, len(permaStorage)+1):
if storage[2] == permaStorage[i].number:
print "This number already exists. No two people can have the same phonenumber!\n"
break
if i == len(permaStorage):
print "hej"
try:
tempbox = Add(storage[1], storage[2])
permaStorage.append(tempbox)
except:
raw_input ("Remember to write name and phonenumber! Press any key to continue \n")
I think problem is that permaStorage is empty list and then u try to:
for i in range(0, len(permaStorage)+1):
if storage[2] == permaStorage[i].number:
will cause an error because permaStorage has 0 items but u trying to get first (i=0, permaStorage[0]) item.
I think you should replace second if clause with first one:
for i in range(0, len(permaStorage)+1):
if i == len(permaStorage):
print "hej"
try:
tempbox = Add(storage[1], storage[2])
permaStorage.append(tempbox)
if storage[2] == permaStorage[i].number:
print "This number already exists. No two people can have the same phonenumber!\n"
break
So in this case if perStorage is blank you will append some value and next if clause will be ok.
Indexing starts at zero in python. Hence, a list of length 5 has the last element index as 4 starting from 0. Change range to range(0, len(permastorage))
You should iterate upto the last element of the list, not beyond.
Try -
for i in range(0, len(permaStorage)):
The list of numbers produced in range() is from the start, but not including the end, so range(3) == [0, 1, 2].
So if your list x has length 10, range(0, len(x)) will give you 0 through 9, which is the correct indices of the elements of your list.
Adding 1 to len(x) will produce the range 0 through 10, and when you try to access x[10], it will fail.
I need to get user input to generate a list of 8 numbers, but when they input a number that is already in the list print and error . Without using the in function to determine if its in the list. Here's what I have so far.
def main():
myList = range(9)
a= True
for i in myList:
while a == True:
usrNum = int(input("Please enter a number: "))
if usrNum != myList[i]:
myList.append(usrNum)
print(i)
main()
Error for above code,
Scripts/untitled4.py", line 18, in main
myList.append(usrNum)
AttributeError: 'range' object has no attribute 'append'
The issue seems to be your way of generating myList. If you generate it with myList = [range(9)] you'll get:
[[0, 1, 2, 3, 4, 5, 6, 7, 8]]
Try using simply:
myList = range(9)
Also, you need to change myList.append[usrNum] with myList.append(usrNum) or you'll get a:
TypeError: 'builtin_function_or_method' object has no attribute '__getitem__'
You could also use wim's suggestion instead of the != operator:
if myList.__contains__(usrNum):
myList.append(usrNum)
There are two ways you can go about this:
Loop through the list to check each element.
The in operator is effectively doing:
for each value in the list:
if the value is what you're looking for
return True
if you reach the end of the list:
return False
If you can add that check into your code, you'll have your problem solved.
Use an alternate way of tracking which elements have been added
Options include a dict, or bits of an int.
For example, create checks = {}. When you add an value to the list, set checks[usrNum] = True. Then checks.get(usrNum, False) will return a boolean indicating whether the number already exists. You can simplify that with a collections.DefaultDict, but I suspect that may be more advanced than you're ready for.
The first is probably the result your instructor is after, so I'll give you a simple version to work with and massage to fit your needs.
myList = []
while True:
usrNum = int(input())
found = False
for v in myList:
if usrNum == v:
found = True
if not found:
myList.append(usrNum)
else:
#number was already in the list, panic!
Most instructors will be more impressed, and hence award better grades, if you can figure out how to do something like method 2, however.
You could do something like this, modify as needed (not sure when/if you want to break when the user enters a number that is already in the list, etc.)
This prompts for user input until they enter an item that already exists in the list, then it prints a message to the user, and stops execution.
def main():
mylist = range(9)
while True:
usrNum = int(input("Please enter a number: "))
if existsinlist(mylist, usrNum):
print("{} is already in the list {}".format(usrNum, mylist))
break
else:
mylist.append(usrNum)
def existsinlist(lst, itm):
for i in lst:
if itm == i:
return True
return False
Perhaps the point of this homework assignment is to help you understand how an operator like in is more efficient to read (and write, and compile) than the explicit loop that I used in the existsinlist function.
Not sure if list-comperehension would be allowable in this case, but you also could've done something like this, without relying on the existsinlist helper function:
def main():
mylist = range(9)
while True:
usrNum = int(input("Please enter a number: "))
if [i for i in mylist if i == usrNum]:
print("{} is already in the list {}".format(usrNum, mylist))
break
else:
mylist.append(usrNum)
In this case, the result of the list-comprehension can be evaluated for truthiness:
An empty list like [] results if no matching value exists, and this will be considered False
A non-empty list will result if at least one matching value exists, and this will be considered True
Yet another option which short-circuits and may be preferable:
if any(usrNum == i for i in mylist)
For example I have a list with a month and the month number, I want to add in another value in [0][1], [1][1],[2][1] every time I input a new value. I'm thinking of the append function however I'm unsure how to re-code that to make that value to input into the the next list along in the list.
list = [[['Jan'],1], [['Feb'],2],[['Mar'],3]]
def month():
global list
count = 0
while value_count < 3:
value = int(input("Enter a value: "))
if value in range(101):
list[0][1].append(value)
count += 1
else:
print('Try Again.')
month()
I want to result with something like:
list = [[['Jan'],1,54], [['Feb'],2,65],[['Mar'],3,62]]
Where 54, 65 and 62 are random numbers a user has entered.
Please see the comments in the code
# list is a function used by python to create lists,
# do not use builtins' names for your variables
my_list = [[['Jan'],1], [['Feb'],2],[['Mar'],3]]
def month():
# print("Don't use globals!\n"*100)
# you can pass arguments to a function, like in "def mont(my_list)"
global my_list
# cycling on a numerical index is complicated, use this syntax
# instead, meaning "use, in turn, the same name to refer to each
# element of the list"
for element in my_list:
value = int(input("Enter a value: "))
# if value in range(101):
if 1: # always true, the test doesn't work as you expect
element.append(value)
else:
print('Try Again.')
month()
print my_list
you can write the input loop like this
value = 101
while not (0<=value<101):
value = int(input(...)
element.append(value)
this way you'll miss the alternative prompt but it's good enough.
I'm trying to implement a function to find occurrences in a list, here's my code:
def all_numbers():
num_list = []
c.execute("SELECT * FROM myTable")
for row in c:
num_list.append(row[1])
return num_list
def compare_results():
look_up_num = raw_input("Lucky number: ")
occurrences = [i for i, x in enumerate(all_numbers()) if x == look_up_num]
return occurrences
I keep getting an empty list instead of the ocurrences even when I enter a number that is on the mentioned list.
Your code does the following:
It fetches everything from the database. Each row is a sequence.
Then, it takes all these results and adds them to a list.
It returns this list.
Next, your code goes through each item list (remember, its a sequence, like a tuple) and fetches the item and its index (this is what enumerate does).
Next, you attempt to compare the sequence with a string, and if it matches, return it as part of a list.
At #5, the script fails because you are comparing a tuple to a string. Here is a simplified example of what you are doing:
>>> def all_numbers():
... return [(1,5), (2,6)]
...
>>> lucky_number = 5
>>> for i, x in enumerate(all_numbers()):
... print('{} {}'.format(i, x))
... if x == lucky_number:
... print 'Found it!'
...
0 (1, 5)
1 (2, 6)
As you can see, at each loop, your x is the tuple, and it will never equal 5; even though actually the row exists.
You can have the database do your dirty work for you, by returning only the number of rows that match your lucky number:
def get_number_count(lucky_number):
""" Returns the number of times the lucky_number
appears in the database """
c.execute('SELECT COUNT(*) FROM myTable WHERE number_column = %s', (lucky_number,))
result = c.fetchone()
return result[0]
def get_input_number():
""" Get the number to be searched in the database """
lookup_num = raw_input('Lucky number: ')
return get_number_count(lookup_num)
raw_input is returning a string. Try converting it to a number.
occurrences = [i for i, x in enumerate(all_numbers()) if x == int(look_up_num)]