I want to iterate over a certain range and create several sets that contain only the current i. (In the code I don't want to do this for every i, but it's about the general principle).
for i in range(5):
s=set(i)
print(s)
It says int object is not iterable. Why doesn't this work? Please keep the answers simple, I'm a newbie.
set() takes a sequence of values to add, and a single integer is not a sequence.
You could wrap it in a tuple or a list:
s = set((i,))
s = set([i])
but the better option is to use the {..} set literal notation:
s = {i}
The notation looks a lot like creating a dictionary, but you only list values, not keys.
Demo (on Python 2, where the representation uses set([..]) notation):
>>> i = 42
>>> set((i,))
set([42])
>>> set([i])
set([42])
>>> {i}
set([42])
Python 3 reflects sets using the literal notation:
>>> i = 42
>>> {i}
{42}
Set constructor set(x) requires x to be some container, like a list, or other set. You pass an integer, python tries to iterate over it, and fails.
In order to do so you need to pass a singleton of x, like that:
for i in range(5):
s = set([i]) # or s = set((i,))
print(s)
or through simplified set constructor
for i in range(5):
s = {i}
print(s)
or construct an empty set and add your element
for i in range(5):
s = set()
s.add(i)
print(s)
instead of using a for loop in range, you can also feed it to set (or a frozenset if you dont need to change it's content):
>>> set(range(5))
# {0, 1, 2, 3, 4}
Related
For example lets say I have a list as below,
list = ['list4','this1','my3','is2'] or [1,6,'one','six']
So now I want to change the index of each element to match the number or make sense as I see fit (needn't be number) like so, (basically change the index of the element to wherever I want)
list = ['this1','is2','my3','list4'] or ['one',1,'six',6]
how do I do this whether there be numbers or not ?
Please help, Thanks in advance.
If you don't wanna use regex and learn it's mini language use this simpler method:
list1 = ['list4','this1', 'he5re', 'my3','is2']
def mySort(string):
if any(char.isdigit() for char in string): #Check if theres a number in the string
return [float(char) for char in string if char.isdigit()][0] #Return list of numbers, and return the first one (we are expecting only one number in the string)
list1.sort(key = mySort)
print(list1)
Inspired by this answer: https://stackoverflow.com/a/4289557/11101156
For the first one, it is easy:
>>> lst = ['list4','this1','my3','is2']
>>> lst = sorted(lst, key=lambda x:int(x[-1]))
>>> lst
['this1', 'is2', 'my3', 'list4']
But this assumes each item is string, and the last character of each item is numeric. Also it works as long as the numeric parts in each item is single digit. Otherwise it breaks. For the second one, you need to define "how you see it fit", in order to sort it in a logic.
If there are multiple numeric characters:
>>> import re
>>> lst = ['lis22t4','th2is21','my3','is2']
>>> sorted(lst, key=lambda x:int(re.search(r'\d+$', x).group(0)))
['is2', 'my3', 'list4', 'this21']
# or,
>>> ['is2', 'my3', 'lis22t4', 'th2is21']
But you can always do:
>>> lst = [1,6,'one','six']
>>> lst = [lst[2], lst[0], lst[3], lst[1]]
>>> lst
['one', 1, 'six', 6]
Also, don't use python built-ins as variable names. list is a bad variable name.
If you just want to move element in position 'y' to position 'x' of a list, you can try this one-liner, using pop and insert:
lst.insert(x, lst.pop(y))
If you know the order how you want to change indexes you can write simple code:
old_list= ['list4','this1','my3','is2']
order = [1, 3, 2, 0]
new_list = [old_list[idx] for idx in order]
If you can write your logic as a function, you can use sorted() and pass your function name as a key:
old_list= ['list4','this1','my3','is2']
def extract_number(string):
digits = ''.join([c for c in string if c.isdigit()])
return int(digits)
new_list = sorted(old_list, key = extract_number)
This case list is sorted by number, which is constructed by combining digits found in a string.
a = [1,2,3,4]
def rep(s, l, ab):
id = l.index(s)
q = s
del(l[id])
l.insert(ab, q)
return l
l = rep(a[0], a, 2)
print(l)
Hope you like this
Its much simpler
I am trying to create a generator function that loops over an iterable sequence while eliminating duplicates and then returns each result in order one at a time (not as a set or list), but I am having difficulty getting it to work. I have found similar questions here, but the responses pretty uniformly result in a list being produced.
I would like the output to be something like:
>>> next(i)
2
>>> next(i)
8
>>> next(i)
4....
I was able to write it as a regular function that produces a list:
def unique(series):
new_series = []
for i in series:
if i not in new_series:
new_series.append(i)
return new_series
series = ([2,8,4,5,5,6,6,6,2,1])
print(unique(series))
I then tried rewriting it as a generator function by eliminating the lines that create a blank list and that append to that list, and then using "yield" instead of "return"; but I’m not getting it to work:
def unique(series):
for i in series:
if i not in new_series:
yield new_series
I don't know if I'm leaving something out or putting too much in. Thank you for any assistance.
Well, to put it simply, you need something to "remember" the values you find. In your first function you were using the new list itself, but in the second one you don't have it, so it fails. You can use a set() for this purpose.
def unique(series):
seen = set()
for i in series:
if i not in seen:
seen.add(i)
yield i
Also, yield should "yield" a single value at once, not the entire new list.
To print out the elements, you'll have to iterate on the generator. Simply doing print(unique([1, 2, 3])) will print the resulting generator object.
>>> print(unique([1, 1, 2, 3]))
<generator object unique at 0x1023bda98>
>>> print(*unique([1, 1, 2, 3]))
1 2 3
>>> for x in unique([1, 1, 2, 3]):
print(x)
1
2
3
Note: * in the second example is the iterable unpack operator.
Try this:
def unique(series):
new_se = []
for i in series:
if i not in new_se:
new_se.append(i)
new_se = list(dict.fromkeys(new_se)) # this will remove duplicates
return new_se
series = [2,8,4,5,5,6,6,6,2,1]
print(unique(series))
I have read this but it does not seem to work for me -
>>> dict = {}
>>> dict[ "a" ] = set(("1111","test1"))
>>> dict
{'a': set(['1111', 'test1'])}
>>> dict[ "a" ].add( ("2222","test2"))
>>> dict
{'a': set(['1111', 'test1', ('2222', 'test2')])}
I was expecting also the first one to look like a tuple - why is that splitted?
I want when doing :
>>> for key,val_set in dict.items() :
... for val in val_set :
... print val
...
1111
test1
('2222', 'test2')
to have 2 items printed not three.
set() takes the values from an iterable and adds each value separately. You passed in a tuple, which is an iterable with two values.
If you wanted to add the tuple itself, you need to wrap that in an iterable, like a list:
set([("1111","test1")])
In Python 3, you can instead use a set display to create a new set:
{("1111","test1")}
A set display takes elements separated by commas, and ("1111", "test1") is one tuple element here.
In the other post, the set was created with a set display (so {tuple1, tuple2, etc.}) and the answers used either an itertools function or a generator expression to produce the sequence of tuples for the set() function to process; a both produce an iterable type.
You have got your answer but I will advice you to use defaultdict instead. All these type of cases like this set([("1111","test1")]) are properly handled. You don't have to worry about initialize too.
Try This.
from collections import defaultdict
s = defaultdict(set)
s["a"].add(("pw", "sq"));
s["a"].add(("wq", "qq"));
s["a"].add(("aa", "bb"));
for k, v in s.items():
print (k)
for g in v:
print (g)
And please note this, I have assumed all your values opposite your keys are sets.
if this is not the case, go with the first one.
When working with a function that returns multiple values with a tuple, I will often find myself using the following idiom to unpack the results from inside a list comprehension.
fiz, buz = zip(*[f(x) for x in input])
Most of the time this works fine, but it throws a ValueError: need more than 0 values to unpack if input is empty. The two ways I can think of to get around this are
fiz = []
buz = []
for x in input:
a, b = f(x)
fiz.append(a)
buz.append(b)
and
if input:
fiz, buz = zip(*[f(x) for x in input])
else:
fiz, buz = [], []
but neither of these feels especially Pythonic—the former is overly verbose and the latter doesn't work if input is a generator rather than a list (in addition to requiring an if/else where I feel like one really shouldn't be needed).
Is there a good simple way to do this? I've mostly been working in Python 2.7 recently, but would also be interested in knowing any Python 3 solutions if they are different.
If f = lambda x: (x,x**2) then this works
x,y = zip(*map(f,input)) if len(input) else ((),())
If input=[], x=() and y=().
If input=[2], x=(2,) and y=(4,)
If input=[2,3], x=(2,3) and y=(4,9)
They're tuples (not lists), but thats thats pretty easy to change.
I would consider using collections.namedtuple() for this sort of thing. I believe the named tuples are deemed more pythonic, and should avoid the need for complicated list comprehensions and zipping / unpacking.
From the documentation:
>>> p = Point(11, y=22) # instantiate with positional or keyword arguments
>>> p[0] + p[1] # indexable like the plain tuple (11, 22)
33
>>> x, y = p # unpack like a regular tuple
>>> x, y
(11, 22)
>>> p.x + p.y # fields also accessible by name
33
>>> p # readable __repr__ with a name=value style
Point(x=11, y=22)
You could use:
fiz = []
buz = []
results = [fiz, buz]
for x in input:
list(map(lambda res, val: res.append(val), results, f(x)))
print(results)
Note about list(map(...)): in Python3, map returns a generator, so we must use it if we want the lambda to be executed.list does it.
(adapted from my answer to Pythonic way to append output of function to several lists, where you could find other ideas.)
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.