Dear public recently I have been receiving an unusual error to do with tuples. I think it has something to do with the spaces. Thanks for the help Adam FYI this program generates all combination and puts numbers together.
Does anybody know whats going on
import intercools
list1 = []
stuff = [1, 2, 3]
for L in range(0, len(stuff+1):
for subset in itertools.combinations(stuff, L):
list1.append(subset)
print(list1)
sep = [map(str,l)for l in list1]
nl = [int(''.join(s)) for s in sep]
print(nl)
There are a number of syntax errors present in your code. To start,
Your outer loop is missing an right parenthesis after stuff
The name of the module is itertools not intercools
After getting through those,
nl = [int(''.join(s)) for s in sep]
ValueError: invalid literal for int() with base 10: ''
So it appears that sep is occasionally empty. It's not clear what your goal is, but it looks like list1 does not contain what you expect.
Related
Problem:
So I was trying to alphabetically sort my list of strings maybe I overlooked something very minor. I have tried both .sort and sorted() but maybe I didn't do it correctly?
Here is my Code:
words = input("Words: ")
list1 = []
list1.append(words.split())
print(sorted(list1))
Expected output-
Input: "a b d c"
Output: ['a', 'b', 'c', 'd']
Current output-
Input: "a b d c"
Output: [['a', 'b', 'd', 'c']]
Your code is not working because you are trying to sort a list inside a list.
When you call words.split() it returns a list. So when you do list1.append(words.split()) it is appending a list into list1.
You should do this:
words = input("Words: ")
list1 = words.split()
print(sorted(list1))
You can try a simple method as follows:
list1 = [i for i in input('Words: ').split(' ')]
print(sorted(list1))
I've tested it. And it is working
Without deviating from your current effort, the only modification you need to do to fix your code is :
words = input("Words: ")
list1 = []
list1.append(words.split())
print(sorted(list1[0]))
Explanation of what you were doing wrong:
The root cause of your confusion is append() .According to python docs,append() takes exactly one argument.
So when you do this,
words.split()
You are trying to append more than 1 element into the list1 and when you append() something more than 1 in a list, it appends as a nested list (i.e a list inside another list.)
To support my explanation you can see that your code fixed by a simple [0]
print(sorted(list1[0]))
That is because your input is stored as a list of list, AND it is stored in the first index (Point to note - 1st index in a python list is 0, hence the usage of list1[0])
Please let me know if I could have explained it in a more simpler way or if you have any other confusions that aid from the above explanation.
This question already has answers here:
How do I remove duplicates from a list, while preserving order?
(30 answers)
Closed 4 years ago.
the program says "TypeError: 'int' object is not iterable"
list=[3,3,2]
print(list)
k=0
for i in list:
for l in list:
if(l>i):
k=l
for j in k:
if(i==j):
del list[i]
print(list)
An easy way to do this is with np.unique.
l=[3,3,2]
print(np.unique(l))
Hope that helps!
Without using any numpy the easiest way I can think of is to start with a new list and then loop through the old list and append the values to the new list that are new. You can cheaply keep track of what has already been used with a set.
def delete_duplicates(old_list):
used = set()
new_list= []
for i in old_list:
if i not in used:
used.add(i)
new_list.append(i)
return new_list
Also, a couple tips on your code. You are getting a TypeError from the for j in k line, it should be for j in range(k). k is just an integer so you can't iterate over it, but range(k) creates an iterable that will do what you want.
Just build another list
>>> list1=[3,2,3]
>>> list2=[]
>>> for i in list1:
... if i in list2:
... pass
... else:
... list2.append(i)
...
>>> list2
[3, 2]
You can always add list1 = list2 at the end if you prefer.
You can use set()
t = [3, 3, 2]
print(t) # prints [3, 3, 2]
t = list(set(t))
print(t) # prints [2, 3]
To remove a duplicate item in a list and get list with unique element, you can always use set() like below:
example:
>>>list1 = [1,1,2,2,3,3,3]
>>>new_unique_list = list(set(list1))
>>> new_unique_list
>>>[1, 2, 3]
You have the following line in your code which produces the error:
for j in k:
k is an int and cannot be iterated over. You probably meant to write for j in list.
There are a couple good answers already. If you really want to write the code yourself however, I'd recommend functional style instead of working in place (i.e. modifying the original array). For example like the following function which is basically a port of Haskell's Data.List.nub.
def nub(list):
'''
Remove duplicate elements from a list.
Singleton lists and empty lists cannot contain duplicates and are therefore returned immediately.
For lists with length gte to two split into head and tail, filter the head from the tail list and then recurse on the filtered list.
'''
if len(list) <= 1: return list
else:
head, *tail = list
return [head] + nub([i for i in tail if i != head])
This is—in my opinion—easier to read and saves you the trouble associated with multiple iteration indexes (since you create a new list).
I am loading a json dataset and assigning it to variables. I tried using it with or without comma but the result is same. Can anyone explain me the significance of this commma here?
with open('datafile01.json', 'rb') as fa, open('datafile02.json', 'rb') as fb:
policies, = json.load(fa).values()
shifts, = json.load(fb).values()
This:
a, = some_sequence
is a degenerate form of sequence unpacking, which only works for sequences of length 1 (else it raises a ValueError with either "too many values to unpack" or "need more than 0 values to unpack").
And this does NOT yield the same result as directly binding the sequence:
>>> d = {"x":1}
>>> a, = d.values()
>>> a
1
>>> a = d.values()
>>> a
dict_values([1])
>>>
i have the task to get the String 'AAAABBBCCDAABBB' into a list like this: ['A','B','C','D','A','B']
I am working on this for 2 hours now, and i can't get the solution. This is my code so far:
list = []
string = 'AAAABBBCCDAABBB'
i = 1
for i in string:
list.append(i)
print(list)
for element in list:
if list[element] == list[element-1]:
list.remove(list[element])
print(list)
I am a newbie to programming, and the error "TypeError: list indices must be integers or slices, not str" always shows up...
I already changed the comparison
if list[element] == list[element-1]
to
if list[element] is list[element-1]
But the error stays the same. I already googled a few times, but there were always lists which didn't need the string-format, but i need it (am i right?).
Thank you for helping!
NoAbL
First of all don't name your variables after built in python statements or data structures like list, tuple or even the name of a module you import, this also applies to files. for example naming your file socket.py and importing the socket module is definitely going to lead to an error (I'll leave you to try that out by yourself)
in your code element is a string, indexes of an iterable must be numbers not strings, so you can tell python
give me the item at position 2.
but right now you're trying to say give me the item at position A and that's not even valid in English, talk-less of a programming language.
you should use the enumerate function if you want to get indexes of an iterable as you loop through it or you could just do
for i in range(len(list))
and loop through the range of the length of the list, you don't really need the elements anyway.
Here is a simpler approach to what you want to do
s = string = 'AAAABBBCCDAABBB'
ls = []
for i in s:
if ls:
if i != ls[-1]:
ls.append(i)
else:
ls.append(i)
print(ls)
It is a different approach, but your problem can be solved using itertools.groupby as follows:
from itertools import groupby
string = 'AAAABBBCCDAABBB'
answer = [group[0] for group in groupby(string)]
print(answer)
Output
['A', 'B', 'C', 'D', 'A', 'B']
According to the documentation, groupby:
Make an iterator that returns consecutive keys and groups from the iterable
In my example we use a list comprehension to iterate over the consecutive keys and groups, and use the index 0 to extract just the key.
You can try the following code:
list = []
string = 'AAAABBBCCDAABBB'
# remove the duplicate character before append to list
prev = ''
for char in string:
if char == prev:
pass
else:
list.append(char)
prev = char
print(list)
Output:
['A', 'B', 'C', 'D', 'A', 'B']
In your loop, element is the string. You want to have the index.
Try for i, element in enumerate(list).
EDIT: i will now be the index of the element you're currently iterating through.
I have written a code which outputs the sub tokens a word. I have defined many cases for which it works perfectly except for one case for which it returns :
['']
instead of :
[]
Is there a way that i can ask python to print the square brackets, without the apostrophe's whenever that case occurs?
Assuming you have a list containing an empty string you can check for it and print the [] symbols (I really don't see a reason for this other than making your code "nicer"(?)):
>>> a = [""]
>>> print(a)
['']
>>> if a==[""]:
print("[]")
[]