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I have a list which includes infinite an unknown number of elements (I don't have control on the length of the list or the type of values)
a = [x, y, z ...]
where x = 1, y = 9, z = 'Hello' ...
And I would like to loop over list "a" and print all the elements' names and values.
I hope if I can implement something like that:
for i in a:
print i $i
I would like to have output as:
x 1
y 9
z Hello
from itertools import cycle # an infinite repeating set
my_infinite = cycle(["a","b","c"])
a=1,b=4,c=99
for val in my_infinite:
print val, globals()[val]
maybe ... its aweful and hackey but it might work
Something like this:
import string
for a, b in zip(string.ascii_lowercase, xrange(1, 27)):
print a, b
You should use a dictionary:
a = {'x': 1, 'y': 2}
for i in a:
print i, a[i]
Related
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I have a working code
task = set()
x = 1
while x != 0:
x = int(input('input your number '))
task.add(x)
print('Just 0 could stop it!')
task.remove(max(task))
print(max(task))
And need to get the same result without using max(). What could be an alternative?
Something like this, unless you have really large sets, I don`t see the advantage
task = set()
x = 1
m = 0
while x != 0:
x = int(input('input your number '))
task.add(x)
if x > m:
m = x
print('Just 0 could stop it!')
task.remove(m)
print(max(task))
Notice this will only work for positive numbers, if you want to the complete int range you should init m like this m = -sys.maxsize - 1
You could use min with a key arg that inverts the element:
>>> task = {1, 2, 3, 4, 5}
>>> max(task)
5
>>> min(task, key=lambda x: -x)
5
Or you could sort it and take the last element...
>>> sorted(task)[-1]
5
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Is there some way to create a list of lists in Python? Analogous to zeroes or similar functions.
What I am thinking of is something like this (pseudocode to follow):
def listoflists(nlists, nitems, fill)
...
listoflists(2,3, -1) # returns [[-1,-1,-1],[-1,-1,-1]]
Not sure if there's a library function to do it, but here's how I would do it:
def listoflists(nList, nItem, fill):
return [[fill for _ in range(nItem)] for _ in range(nList)]
This uses list comprehension to make a list of size nItem nList times.
You can create a list
root_list = []
and just append your sublists
root_list.append([1, 1, 1])
just need to do this:
def list_of_list(n_list, n_elements, fill):
x = [fill] * n_elements
y = [x] * n_list
return y
print(list_of_list(2, 3, -1))
Bye.
This should work, thanks!
def listoflists(nlists, nitems, fill):
f_list = []
for i in range(nlists):
n_list = []
for j in range(nitems):
n_list.append(fill)
f_list.append(n_list)
n_list = []
return f_list
print(listoflists(2,3,-1))
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I play to HackNet game and i have to guess a word to bypass a firewall.
The key makes 6 characters long and contains the letters K,K,K,U,A,N.
What is the simplest way to generate all possible combinations either in bash or in python ? (bonus point for bash)
Here is a backtracking-based solution in Python:
db = {"K" : 3, "U" : 1, "A" : 1, "N" : 1}
N = 6
def possibilities(v):
l = []
for k in db.keys():
if v.count(k) < db[k]:
l.append(k)
return l
def generateImpl(a):
if len(a) < N:
lst = []
for c in possibilities(a):
lst += generateImpl(a+[c])
return lst
else:
return [''.join(a)]
def generate():
return generateImpl([])
Just run generate() to obtain a list of possible words.
You have 6 letters and need to find a combination of 6 letters. If you are not using the same character in ['K','K','K','U','A','N'] again and again, then there is only 1 permutation.
If you can use the same character again and again, you can use the following code to generate all possible combinations.
import itertools
y = itertools.combinations_with_replacement(['K','U','A','N'],6)
for i in list(y):
print(i)
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I have tried to use the max code but it printed out the largest number not the variable name.
a=3
b=4
c=7
d=6
e=8
max(a,b,c,d,e)
this is the code I've tried but it printed 8 instead of e.
forgot to add that my values for a b c etc came from a loop so im not suppose to know their values myself...
You need to use a dictionary or a named tuple to do something like that.
>>> my_list = {'a': 1, 'b': 2, 'c': 3}
>>> max(my_list, key=my_list.get)
'c'
Let me know if there are any other methods.
Try this
dct={'a':3 ,'b':4 ,'c':7 ,'d':6 ,'e':8}
for i, j in dct.iteritems():
if j == max(dct.values()):
print i
From this question, there seems to be a way to retrieve the name of your variable (see caveat below before using):
a = 3
b = 4
c = 7
d = 6
e = 8
m = max(a, b, c, d, e)
for k, v in list(locals().items()):
if v is m:
v_as_str = k
break
v_as_str
output:
'e'
Caveat:
This may work if you can be sure that there are no other (unrelated) variables with the same value as the max of a to e
somewhere in your code, and appearing earlier in the locals.
Thanks to #D.Everhard & #asongtoruin in the comments
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List1 =['000095', '000094', '000092', '000101', '000099', '000096', '000095']
def makecycle(list, startElement):
A loop which forms the bottom list which is made from the upper one's elements!
if i pass that function the start element and the list it shoul print like this:
makecycle(list1, 000094) it should print:
['000094', '000092', '000101', '000099', '000096', '000095', '000094']
and if pass
makecycle(list1, 000101) it should print:
['000101', '000099', '000096', '000095', '000094', '000092', '000101']
and if pass
makecycle(list1, 000092) it should print:
['000092', '000101', '000099', '000096', '000095', '000094', '000092']
i know its kinda not clear enough but thats all i can point!
def makecycle(list1,startElement):
ind = list1.index(startElement)
l = len(list1)
i = ind
list2 = []
while ind < (l-1):
list2.append(list1[ind])
ind = ind + 1
for x in range(i):
if list1[x] not in list2:
list2.append(list1[x])
print(list2)