I have a list, which potentially contains a list (and this subsequently can also can be made up of lists, ad infinitum). Is there a way to test if any element at the 'bottom' of these lists has an integer? The list will always have numbers (floats or ints) eventually.
As I don't know the number of nested lists of it beforehand, the only way I can think of doing so would be something like,
x = [[[[5]]]]
if (len(str(x)) != len(str(x).replace('.','')) or ('int' not in str(x)):
int_in_list = False
Is there a more logical way of doing this?
I'd recommend using something like collapse() from the more-itertools library, which will recursively descend into lists and other iterable types and yield the fundamental non-iterable elements. You can then just check whether any of the elements yielded by collapse() is an integer.
any(isinstance(x, int) for x in collapse(...))
Here ya go:
numlist = [[[[[1]]]]]
def recursive_flatten(lst):
for item in lst:
if isinstance(item, list):
yield from recursive_flatten(item)
yield item
if any(isinstance(item, int) for item in recursive_flatten(numlist)):
# Horray
pass
Related
For nested lists, what would the pseudocode look like so that
mylist = [[1,2],[3,[4,5]],6]
def mydeepcopy(mylist)
>> [[1,2],[3,[4,5]],6]
where it returns a completely different list through all layers
here is all I can think of
def mycopy(mylist):
if mylist is not 1-dimensional:
mycopy(elem for elem in mylist)
else:
add mylist.copy() to new list
You have to collect the results of the recursion and return them.
def mycopy(mylist):
newlist = []
for element in mylist:
if isinstance(element, list):
newlist.append(mycopy(element))
else:
newlist.append(element)
return newlist
This is overly simplified, as it only deals with lists, not tuples and dictionaries.
Note that "is not 1-dimensional" is not something you test for the entire list. It can contain a mixture of nested lists and other objects, so you need to test each element in the loop.
New to Python here. I've figured out that it's rather trivial to check if a list has all unique elements using
if len(lst) > len(set(lst))
But how would I deal with a case like this?
all_different([[],[],[],[]]) #False
I can't use the set() function in this example. I understand how I might do this (rather inefficiently) with nested for loops in Java. But not sure how I could implement this in Python.
You can't hash lists as they're mutable, but you can convert them to tuples, and hash them instead. Like so: set(tuple(elem) for elem in lst)
Your comparison would then become:
myset = set(tuple(elem) for elem in lst)
if len(lst) > len(myset)
I have a long list that is made up of many tuples (over 100)that all contain 3 items that are strings
first_list = ('the','cat','went'),('back','too','scho'),('t/e','s/e/t','o/ve') etc
Many of the tuples are identically so i am using the set function to get out a unique set
Long_list = set(first_list)
i need the list in its original format , but i also need a duplicate list where the data has been cleaned
I need to remove all the "/" and Replace them with "#"
i can t seem to do this process. Initially i tried creating a foor loop to go through my list and then carry out the find and replace method.
The way i have done it gives me a new list that is made up of items , so the sets of tuples are not retained
for small_tuple in Long_list:
the_list = list(small_tuple)
for uncleaned_string in the_list:
time = uncleaned_string.replace('/','#')
last_list.append(time)
print last_list
Is there a way i can retain my original format of 3 items within tuple when i convert it back ?
tuple(myList) will convert myList into a tuple, provided that myList is something iterable like a list, a tuple, or a generator.
To convert a lists of lists in a list of tuples, using a list comprehension expression:
last_list = [tuple(x) for x in Long_list]
or, to also perform your string replacement:
last_list = [tuple(y.replace('/', '#') for y in x) for x in Long_list]
From Python's reference:
tuple( [iterable] )
Return a tuple whose items are the same and in the same order as iterable‘s items. iterable may be a sequence, a container that supports iteration, or an iterator object. If iterable is already a tuple, it is returned unchanged. For instance, tuple('abc') returns ('a', 'b', 'c') and tuple([1, 2, 3]) returns (1, 2, 3). If no argument is given, returns a new empty tuple, ().
tuple is an immutable sequence type, as documented in Sequence Types — str, unicode, list, tuple, bytearray, buffer, xrange. For other containers see the built in dict, list and [set] classes, and the collections module.
You could do something like the following using a list comprehension converted into a tuple:
for small_tuple in Long_list:
the_list = list(small_tuple)
last_list.append(tuple([uncleaned_string.replace('/','#') for uncleaned_string in the_list]))
print last_list
last_list = [tuple(s.replace('/', '#') for s in t) for t in Long_list]
Modified your code to fit your need
for small_tuple in Long_list:
the_list = list(small_tuple)
res=[]
for uncleaned_string in the_list:
time = uncleaned_string.replace('/','#')
res.append(time)
last_list.append(tuple(res))
print last_list
You can use an OrderedDict to remove the duplicates while preserving the order.
from collections import OrderedDict
x = OrderedDict.fromkeys(first_list)
Long_list = list(x)
Long_list contains unique tuples with order same as of first_list.
I have a problem:
list = [1,2,3,4,5]
a= 3
if a==[item for item in list]:
print(sth)
why the program never print?
thanks...
You're comparing an integer to a list, which will never return True as they are different types. Note that [item for item in list] is exactly the same as just saying list.
You're probably wondering if 3 is in the list; so you can do:
if a in list:
print(sth)
Or even:
if any(a == item for item in list):
print(sth)
(Although you really should just use the first option. I only put the second option in as it looks similar to your example :p)
As a side note, you shouldn't be naming lists list, or dictionaries dict, as they are built-in types already, and you're just overriding them :p.
A multidimensional list like l=[[1,2],[3,4]] could be converted to a 1D one by doing sum(l,[]). How does this happen?
(This doesn't work directly for higher multidimensional lists, but it can be repeated to handle those cases. For example if A is a 3D-list, then sum(sum(A),[]),[]) will flatten A to a 1D list.)
If your list nested is, as you say, "2D" (meaning that you only want to go one level down, and all 1-level-down items of nested are lists), a simple list comprehension:
flat = [x for sublist in nested for x in sublist]
is the approach I'd recommend -- much more efficient than summing would be (sum is intended for numbers -- it was just too much of a bother to somehow make it block all attempts to "sum" non-numbers... I was the original proposer and first implementer of sum in the Python standard library, so I guess I should know;-).
If you want to go down "as deep as it takes" (for deeply nested lists), recursion is the simplest way, although by eliminating the recursion you can get higher performance (at the price of higher complication).
This recipe suggests a recursive solution, a recursion elimination, and other approaches
(all instructive, though none as simple as the one-liner I suggested earlier in this answer).
sum adds a sequence together using the + operator. e.g sum([1,2,3]) == 6. The 2nd parameter is an optional start value which defaults to 0. e.g. sum([1,2,3], 10) == 16.
In your example it does [] + [1,2] + [3,4] where + on 2 lists concatenates them together. Therefore the result is [1,2,3,4]
The empty list is required as the 2nd paramter to sum because, as mentioned above, the default is for sum to add to 0 (i.e. 0 + [1,2] + [3,4]) which would result in unsupported operand type(s) for +: 'int' and 'list'
This is the relevant section of the help for sum:
sum(sequence[, start]) -> value
Returns the sum of a sequence of
numbers (NOT strings) plus the value
of parameter 'start' (which defaults
to 0).
Note
As wallacoloo comented this is not a general solution for flattening any multi dimensional list. It just works for a list of 1D lists due to the behavior described above.
Update
For a way to flatten 1 level of nesting see this recipe from the itertools page:
def flatten(listOfLists):
"Flatten one level of nesting"
return chain.from_iterable(listOfLists)
To flatten more deeply nested lists (including irregularly nested lists) see the accepted answer to this question (there are also some other questions linked to from that question itself.)
Note that the recipe returns an itertools.chain object (which is iterable) and the other question's answer returns a generator object so you need to wrap either of these in a call to list if you want the full list rather than iterating over it. e.g. list(flatten(my_list_of_lists)).
For any kind of multidiamentional array, this code will do flattening to one dimension :
def flatten(l):
try:
return flatten(l[0]) + (flatten(l[1:]) if len(l) > 1 else []) if type(l) is list else [l]
except IndexError:
return []
It looks to me more like you're looking for a final answer of:
[3, 7]
For that you're best off with a list comprehension
>>> l=[[1,2],[3,4]]
>>> [x+y for x,y in l]
[3, 7]
I wrote a program to do multi-dimensional flattening using recursion. If anyone has comments on making the program better, you can always see me smiling:
def flatten(l):
lf=[]
li=[]
ll=[]
p=0
for i in l:
if type(i).__name__=='list':
li.append(i)
else:
lf.append(i)
ll=[x for i in li for x in i]
lf.extend(ll)
for i in lf:
if type(i).__name__ =='list':
#not completely flattened
flatten(lf)
else:
p=p+1
continue
if p==len(lf):
print(lf)
I've written this function:
def make_array_single_dimension(l):
l2 = []
for x in l:
if type(x).__name__ == "list":
l2 += make_array_single_dimension(x)
else:
l2.append(x)
return l2
It works as well!
The + operator concatenates lists and the starting value is [] an empty list.