code A :
t1 = {}
t1[0] = -5
t1[1] = 10.5
code B :
t2 = []
t2[0] = -5
t2[1] = 10.5
why code B has "IndexError: list assignment index out of range" and how to solve it?
Dictionaries are hashsets. They pair arbitrary (hashable) keys with values, and there is no expectation that those keys are consecutive, or even in fact numbers
replacement_dict = {'old_name': 'new_name'} # you could use this to implement a find/replace
By comparison, lists are densely-packed and their indices (not keys -- this is a different term accessed the same way with the object[index] notation) are numbers. Because of this, you can't just access a random value that's greater than the length of the list, you must use append instead.
lst = []
lst[0] = 'blah' # throws an IndexError because len(lst) is 0
lst.append('blah')
assert lst[0] == 'blah
Dictionaries work like key-value pairs. Every time you assign a new value in a dictionary, you create a new key-value pair.
A list is like an array that you can extend at will, but if you try to access an index that is over its current size, it will return an error. You typically extend a list by using t2.append(value).
A dictionary allows assignment of elements that don't exist yet. Lists don't. That's just the way they're designed.
You can fix this two ways. First is to initialize the list to the size it needs to be:
t2 = [None]*2
The second is to call append instead of using =:
t2 = []
t2.append(-5)
t2.append(10.5)
A dictionary stores data with name values.
dictionary = {"name": "value"}
print(dictionary["name"])
# Prints "value".
A list stores a sequence of values. There aren't any names for the values, they are accessed via an index.
list = ["value", "other value"]
print(list[0])
# Prints "value".
To fix your problem use append.
t2 = []
t2.append(-5)
t2.append(10.5)
Related
I'm trying to simply append to a list of lists but cannot find a clean example of how to do that. I've looked at dozens of examples but they are all for appending to a one-dimensional list or extending lists.
Sample code:
testList = []
print(testList)
testList.append(3000)
print(testList)
testList[3000].append(1)
testList[3000].append(2)
print(testList)
Expected result:
testList[3000][1, 2]
Actual result:
[]
[3000]
Traceback (most recent call last):
File ".\test.py", line 5, in <module>
testList[3000].append(1)
IndexError: list index out of range
The first problem I see is that when you call testList.append() you are including the [3000]. That is problematic because with a list, that syntax means you're looking for the element at index 3000 within testList. All you need to do is call testList.append(<thing_to_append>) to append an item to testList.
The other problem is that you're expecting [1, 2] to be a separate list, but instead you're appending them each to testList.
If you want testList to be composed of multiple lists, a good starting point would be to instantiate them individually, and to subsequently append to testList. This should help you conceptualize the nested structure you're looking for.
For example:
testList = []
three_k = [3000]
one_two = [1, 2]
testList.append(three_k)
testList.append(one_two)
print(testList)
From there, the you can actually use the indexes of the nested lists to append to them. So if [3000] is the list at index 0 (zero), you can append to it by doing: testList[0].append(<new_append_thing>).
First, thank you to everyone for the quick responses. Several of you got me thinking in the right direction but my original question wasn't complete (apologies) so I'm adding more context here that's hopefully helpful for someone else in the future.
wp-overwatch.com jogged my memory and I realized that after working with only dictionaries in my application for days, I was treating the "3000" like a dictionary key instead of the list index. ("3000" is an example of an ID number that I have to use to track one of the lists of numbers.)
I couldn't use a dictionary, however, because I need to add new entries, remove the first entry, and calculate average for the numbers I'm working with. The answer was to create a dictionary of lists.
Example test code I used:
testDict = {}
blah10 = 10
blah20 = 20
blah30 = 30
blah40 = 40
exampleId = 3000
if exampleId == 3000:
testDict[3000] = []
testDict[3000].append(blah10)
testDict[3000].append(blah20)
print(testDict)
testDict[3000].pop(0) # Remove first entry
print(testDict)
testDict[3000].append(blah30) # Add new number to the list
average = sum(testDict[3000]) / len(testDict[3000])
print(average)
if exampleId == 3001:
testDict[3001].append(blah30)
testDict[3001].append(blah40)
Result:
{3000: [10, 20]}
{3000: [20]}
25.0
testList[3000].append(1) is telling Python to grab the 3000th item in the list and call the append function on it. Since you don't have 3000 items in the list, that's why you're getting that error.
If you want to lookup an item by a value such as 3000 instead of by its position in the list, then what you're wanting is not a list but a dictionary.
Using a dictionary you can do:
>>> testList = {} # use curly brackets for a dictionary
>>> print(testList)
{}
>>> testList[3000] = [] # create a new item with the lookup key of 3000 and set the corresponding value to an empty list
>>> print(testList)
{3000: []}
>>> testList[3000].append(1)
>>> testList[3000].append(2)
>>> print(testList)
{3000: [1, 2]}
>>>
It is because according to your program the python interpreter will look for the 3000 index at the list and try to append the given number in the index of 3000 but there is not that number so it will print error.
To fix it:
testList = []
print(testList)
testList.append(3000)
print(testList)
testList.append([1])
testList[1].append(2)
print(testList)
Using the index you can append the value as I appended.
list.append adds an object to the end of a list. So doing,
listA = []
listA.append(1)
now listA will have only the object 1 like [1].
you can construct a bigger list doing the following
listA = [1]*3000
which will give you a list of 3000 times 1 [1,1,1,1,1,...].
If you want to contract a c-like array you should do the following
listA = [[1]*3000 for _ in range(3000)]
Now you have a list of lists like [[1,1,1,1,....], [1,1,1,....], .... ]
PS Be very careful using [1]*3000 which in this case works
but in case you use it with a variable it will have side effects. For example,
a = 1
listA = [a]*3000
gives you the same result but if any of the variables 'a' change then all will be changed the same way.
The most safe is, using list comprehensions
a = 1
listA = [a for _ in range(3000)]
In order to update any value, just do the following,
listA[656] = 999
Lastly, in order to extend the above list just type
listA.append(999)
and then at the index 3000 (starting from zero) you will find 999
import random
dictionary = {'dog': 1,'cat': 2,'animal': 3,'horse': 4}
keys = random.shuffle(list(dictionary.keys())*3)
values = list(dictionary.values())*3
random_key = []
random_key_value = []
random_key.append(keys.pop())
random_key_value.append(???)
For random_key_values.append, I need to add the value that corresponds to the key that was popped. How can I achieve this? I need to make use of multiples of the list and I can't multiply a dictionary directly, either.
I'm going on python (you should specify the language in your question).
If I understand, you want to multiply the elements in the dictionary. So
list(dictionary.keys()) * 3
is not your solution: [1,2] * 3 results in [1,2,1,2,1,2]
Try instead list comprehension:
[i * 3 for i in dictionary.keys()]
To take into account the order (because you shuffle it) shuffle the keys before the multiplication, then create the values list (in the same order that the shuffled keys) and finally multiply the keys:
keys = dictionary.keys()
random.shuffle(keys)
values = [dictionary[i]*3 for i in keys]
keys = [i * 3 for i in keys]
And finally:
random_key.append(keys.pop())
random_key_value.append(values.pop())
Also take care about the random function, it doesn't work as you are using it. See the documentation.
I have a list in Python with certain elements. I want to replace these elements with their corresponding elements from another list.
I want to have another list that relates elements in a list like:
x = ['red','orange','yellow','green','blue','purple','pink']
y = ['cherry','orange','banana','apple','blueberry','eggplant','grapefruit']
so that a new list will be created with the the corresponding elements from y whose elements are in x. So
r = ['green','red','purple']
will become
r = ['apple','cherry','eggplant']
Thanks
First create a mapping from one list to another:
my_mapping = dict(zip(x, y))
The zip part is mainly a formality: dict expects the argument to be a sequence of pairs, rather than a pair of sequences.
Then apply this mapping to every member of r using a list comprehension:
new_r = [my_mapping[elem] for elem in r]
Dictionaries are the best option for this. They are maps of key-value pairs so that if I index a dictionary with a key, it will return the value under that key.
Let's make a new dictionary using the zip function that takes two lists:
mapper = dict(zip(x,y))
Now, if we want to change a list r to its counterpart with elements from the new list:
r = [mapper[i] for i in r]
This takes each element in r and uses the dictionary to turn it into its counterpart.
I have this list:
source = ['sourceid', 'SubSourcePontiflex', 'acq_source', 'OptInSource', 'source',
'SourceID', 'Sub-Source', 'SubSource', 'LeadSource_295', 'Source',
'SourceCode', 'source_code', 'SourceSubID']
I am iterating over XML in python to create a dictionary for each child node. The dictionary varies in length and keys with each iteration. Sometimes the dictionary will contain a key that is also an item in this list. Sometimes it wont. What I want to be able to do is, if a key in the dictionary is also an item in this list then append the value to a new list. If none of the keys in the dictionary are in list source, I'd like to append a default value. I'm really having a brain block on how to do this. Any help would be appreciated.
Just use the in keyword to check for membership of some key in a dictionary.
The following example will print [3, 1] since 3 and 1 are keys in the dictionary and also elements of the list.
someList = [8, 9, 7, 3, 1]
someDict = {1:2, 2:3, 3:4, 4:5, 5:6}
intersection = [i for i in someList if i in someDict]
print(intersection)
You can just check if this intersection list is empty at every iteration. If the list is empty then you know that no items in the list are keys in the dictionary.
in_source_and_dict = set(mydict.keys()).intersection(set(source))
in_dict_not_source = set(mydict.keys()) - set(source)
in_source_not_dict = set(source) - set(mydict.keys())
Iterate over the result of which one you want. In this case I guess you'll want to iterate over in_source_not_dict to provide default values.
In Python 3, you can perform set operations directly on the object returned by dict.keys():
in_source_and_dict = mydict.keys() & source
in_dict_not_source = mydict.keys() - source
in_source_not_dict = source - mydict.keys()
This will also work in Python 2.7 if you replace .keys() by .viewkeys().
my_dict = { some values }
values = []
for s in sources:
if my_dict.get(s):
values += [s]
if not values:
values += [default]
You can loop through the sources array and see if there is a value for that source in the dictionary. If there is, append it to values. After that loop, if values is empty, append the default vaule.
Note, if you have a key, value pair in your dictionary (val, None) then you will not append the None value to the end of the list. If that is an issue you will probably not want to use this solution.
You can do this with the any() function
dict = {...}
keys = [...]
if not any(key in dict for key in keys):
# no keys here
Equivalently, with all() (DeMorgan's laws):
if all(key not in dict for key in keys):
# no keys here
This is an incredibly simple question (I'm new to Python).
I basically want a data structure like a PHP array -- i.e., I want to initialise it and then just add values into it.
As far as I can tell, this is not possible with Python, so I've got the maximum value I might want to use as an index, but I can't figure out how to create an empty list of a specified length.
Also, is a list the right data structure to use to model what feels like it should just be an array? I tried to use an array, but it seemed unhappy with storing strings.
Edit: Sorry, I didn't explain very clearly what I was looking for. When I add items into the list, I do not want to put them in in sequence, but rather I want to insert them into specified slots in the list.
I.e., I want to be able to do this:
list = []
for row in rows:
c = list_of_categories.index(row["id"])
print c
list[c] = row["name"]
Depending on how you are going to use the list, it may be that you actually want a dictionary. This will work:
d = {}
for row in rows:
c = list_of_categories.index(row["id"])
print c
d[c] = row["name"]
... or more compactly:
d = dict((list_of_categories.index(row['id']), row['name']) for row in rows)
print d
PHP arrays are much more like Python dicts than they are like Python lists. For example, they can have strings for keys.
And confusingly, Python has an array module, which is described as "efficient arrays of numeric values", which is definitely not what you want.
If the number of items you want is known in advance, and you want to access them using integer, 0-based, consecutive indices, you might try this:
n = 3
array = n * [None]
print array
array[2] = 11
array[1] = 47
array[0] = 42
print array
This prints:
[None, None, None]
[42, 47, 11]
Use the list constructor, and append your items, like this:
l = list ()
l.append ("foo")
l.append (3)
print (l)
gives me ['foo', 3], which should be what you want. See the documentation on list and the sequence type documentation.
EDIT Updated
For inserting, use insert, like this:
l = list ()
l.append ("foo")
l.append (3)
l.insert (1, "new")
print (l)
which prints ['foo', 'new', 3]
http://diveintopython3.ep.io/native-datatypes.html#lists
You don't need to create empty lists with a specified length. You just add to them and query about their current length if needed.
What you can't do without preparing to catch an exception is to use a non existent index. Which is probably what you are used to in PHP.
You can use this syntax to create a list with n elements:
lst = [0] * n
But be careful! The list will contain n copies of this object. If this object is mutable and you change one element, then all copies will be changed! In this case you should use:
lst = [some_object() for i in xrange(n)]
Then you can access these elements:
for i in xrange(n):
lst[i] += 1
A Python list is comparable to a vector in other languages. It is a resizable array, not a linked list.
Sounds like what you need might be a dictionary rather than an array if you want to insert into specified indices.
dict = {'a': 1, 'b': 2, 'c': 3}
dict['a']
1
I agree with ned that you probably need a dictionary for what you're trying to do. But here's a way to get a list of those lists of categories you can do this:
lst = [list_of_categories.index(row["id"]) for row in rows]
use a dictionary, because what you're really asking for is a structure you can access by arbitrary keys
list = {}
for row in rows:
c = list_of_categories.index(row["id"])
print c
list[c] = row["name"]
Then you can iterate through the known contents with:
for x in list.values():
print x
Or check if something exists in the "list":
if 3 in list:
print "it's there"
I'm not sure if I understood what you mean or want to do, but it seems that you want a list which
is dictonary-like where the index is the key. Even if I think, the usage of a dictonary would be a better
choice, here's my answer: Got a problem - make an object:
class MyList(UserList.UserList):
NO_ITEM = 'noitem'
def insertAt(self, item, index):
length = len(self)
if index < length:
self[index] = item
elif index == length:
self.append(item)
else:
for i in range(0, index-length):
self.append(self.NO_ITEM)
self.append(item)
Maybe some errors in the python syntax (didn't check), but in principle it should work.
Of course the else case works also for the elif, but I thought, it might be a little harder
to read this way.