By user1877442


2013-02-27 12:26:23 8 Comments

I'm trying to make a function that will compare multiple variables to an integer and output a string of three letters. I was wondering if there was a way to translate this into Python. So say:

x = 0
y = 1
z = 3
mylist = []

if x or y or z == 0 :
    mylist.append("c")
if x or y or z == 1 :
    mylist.append("d")
if x or y or z == 2 :
    mylist.append("e")
if x or y or z == 3 : 
    mylist.append("f")

which would return a list of

["c", "d", "f"]

Is something like this possible?

20 comments

@Sergei 2019-02-17 17:55:26

Maybe you need direct formula for output bits set.

x=0 or y=0 or z=0 is equivalent to xyz = 0

x=1 or y=1 or z=1 is equivalent to (x-1)(y-1)(z-1)=0

x=2 or y=2 or z=2 is equivalent to (x-2)(y-2)(z-2)=0

lets map to bits: 'c':1 'd':0xb10 'e':0xb100 'f':0xb1000

relation of isc:

if xyz=0 then isc=1 else isc=0

use math if formula https://youtu.be/KAdKCgBGK0k?list=PLnI9xbPdZUAmUL8htSl6vToPQRRN3hhFp&t=315

[c]: (xyz=0 and isc=1) or (((xyz=0 and isc=1) or (isc=0)) and (isc=0))

[d]: ((x-1)(y-1)(z-1)=0 and isc=2) or (((xyz=0 and isd=2) or (isc=0)) and (isc=0))

...

connect these formulas by following logic:

logic and is sum of squares of equations

logic or is product of equations

and you'll have total equation

express sum and you have total formula of sum

then sum&1 is c, sum&2 is d, sum&4 is e, sum&5 is f

after this you may form predefined array where index of string element would correspond to ready string.

array[sum] gives you the string.

@Siddharth Satpathy 2018-12-03 05:13:18

You can try the method shown below. In this method, you will have the freedom to specify/input the number of variables that you wish to enter.

mydict = {0:"c", 1:"d", 2:"e", 3:"f"}
mylist= []

num_var = int(raw_input("How many variables? ")) #Enter 3 when asked for input.

for i in range(num_var): 
    ''' Enter 0 as first input, 1 as second input and 3 as third input.'''
    globals()['var'+str('i').zfill(3)] = int(raw_input("Enter an integer between 0 and 3 "))
    mylist += mydict[globals()['var'+str('i').zfill(3)]]

print mylist
>>> ['c', 'd', 'f']

@Seenivasan 2018-11-06 14:26:24

It can be done easily as

for value in [var1,var2,var3]:
     li.append("targetValue")

@Md. Alamin Mahamud 2018-09-20 02:18:55

To test multiple variables with one single value: if 1 in {a,b,c}:

To test multiple values with one variable: if a in {1, 2, 3}:

@Rolf of Saxony 2016-09-10 15:44:32

All of the excellent answers provided here concentrate on the specific requirement of the original poster and concentrate on the if 1 in {x,y,z} solution put forward by Martijn Pieters.
What they ignore is the broader implication of the question:
How do I test one variable against multiple values?
The solution provided will not work for partial hits if using strings for example:
Test if the string "Wild" is in multiple values

>>> x = "Wild things"
>>> y = "throttle it back"
>>> z = "in the beginning"
>>> if "Wild" in {x, y, z}: print (True)
... 

or

>>> x = "Wild things"
>>> y = "throttle it back"
>>> z = "in the beginning"
>>> if "Wild" in [x, y, z]: print (True)
... 

for this scenario it's easiest to convert to a string

>>> [x, y, z]
['Wild things', 'throttle it back', 'in the beginning']
>>> {x, y, z}
{'in the beginning', 'throttle it back', 'Wild things'}
>>> 

>>> if "Wild" in str([x, y, z]): print (True)
... 
True
>>> if "Wild" in str({x, y, z}): print (True)
... 
True

It should be noted however, as mentioned by @codeforester, that word boundries are lost with this method, as in:

>>> x=['Wild things', 'throttle it back', 'in the beginning']
>>> if "rot" in str(x): print(True)
... 
True

the 3 letters rot do exist in combination in the list but not as an individual word. Testing for " rot " would fail but if one of the list items were "rot in hell", that would fail as well.
The upshot being, be careful with your search criteria if using this method and be aware that it does have this limitation.

@codeforester 2018-06-21 21:08:18

With the "convert to a string" approach, we lose the word boundaries and the in check may not do the right thing.

@Rolf of Saxony 2018-06-22 07:41:49

@codeforester Your point is a fair one and I'll mention it in the answer.

@paarsa 2015-04-10 03:29:39

If you want to use if, else statements following is another solution:

myList = []
aList = [0, 1, 3]

for l in aList:
    if l==0: myList.append('c')
    elif l==1: myList.append('d')
    elif l==2: myList.append('e')
    elif l==3: myList.append('f')

print(myList)

@firelynx 2018-08-27 09:45:00

Looks like you're building some kind of Caesar cipher.

A much more generalized approach is this:

input_values = (0, 1, 3)
origo = ord('c')
[chr(val + origo) for val in inputs]

outputs

['c', 'd', 'f']

Not sure if it's a desired side effect of your code, but the order of your output will always be sorted.

If this is what you want, the final line can be changed to:

sorted([chr(val + origo) for val in inputs])

@rsalmei 2018-08-06 06:26:34

The most mnemonic way of representing your pseudo-code in Python would be:

x = 0
y = 1
z = 3
mylist = []

if any(v == 0 for v in (x, y, z)):
    mylist.append("c")
if any(v == 1 for v in (x, y, z)):
    mylist.append("d")
if any(v == 2 for v in (x, y, z)):
    mylist.append("e")
if any(v == 3 for v in (x, y, z)):
    mylist.append("f")

@Rohit Gawas 2018-07-31 16:54:00

You can use dictionary :

x = 0
y = 1
z = 3
list=[]
dict = {0: 'c', 1: 'd', 2: 'e', 3: 'f'}
if x in dict:
    list.append(dict[x])
else:
    pass

if y in dict:
    list.append(dict[y])
else:
    pass
if z in dict:
    list.append(dict[z])
else:
    pass

print list

@Sergei 2019-02-19 04:49:25

This may append same more then once this. Set?

@B. M. 2015-06-09 20:46:34

Set is the good approach here, because it orders the variables, what seems to be your goal here. {z,y,x} is {0,1,3} whatever the order of the parameters.

>>> ["cdef"[i] for i in {z,x,y}]
['c', 'd', 'f']

This way, the whole solution is O(n).

@Raniz 2015-06-10 04:19:22

You should add a description of what your code accomplishes and how it does it. Short answers using only code is discouraged

@Martijn Pieters 2013-02-27 12:27:40

You misunderstand how boolean expressions work; they don't work like an English sentence and guess that you are talking about the same comparison for all names here. You are looking for:

if x == 1 or y == 1 or z == 1:

x and y are otherwise evaluated on their own (False if 0, True otherwise).

You can shorten that using a containment test against a tuple:

if 1 in (x, y, z):

or better still:

if 1 in {x, y, z}:

using a set to take advantage of the constant-cost membership test (in takes a fixed amount of time whatever the left-hand operand is).

When you use or, python sees each side of the operator as separate expressions. The expression x or y == 1 is treated as first a boolean test for x, then if that is False, the expression y == 1 is tested.

This is due to operator precedence. The or operator has a lower precedence than the == test, so the latter is evaluated first.

However, even if this were not the case, and the expression x or y or z == 1 was actually interpreted as (x or y or z) == 1 instead, this would still not do what you expect it to do.

x or y or z would evaluate to the first argument that is 'truthy', e.g. not False, numeric 0 or empty (see boolean expressions for details on what Python considers false in a boolean context).

So for the values x = 2; y = 1; z = 0, x or y or z would resolve to 2, because that is the first true-like value in the arguments. Then 2 == 1 would be False, even though y == 1 would be True.

The same would apply to the inverse; testing multiple values against a single variable; x == 1 or 2 or 3 would fail for the same reasons. Use x == 1 or x == 2 or x == 3 or x in {1, 2, 3}.

@Private 2013-02-27 13:14:59

can you explain a bit more why the OP misunderstands booleans?

@SingleNegationElimination 2013-10-24 15:27:12

I wouldn't be so quick to go for the set version. Tuple's are very cheap to create and iterate over. On my machine at least, tuples are faster than sets so long as the size of the tuple is around 4-8 elements. If you have to scan more than that, use a set, but if you are looking for an item out of 2-4 possibilities, a tuple is still faster! If you can arrange for the most likely case to be first in the tuple, the win is even bigger: (my test: timeit.timeit('0 in {seq}'.format(seq=tuple(range(9, -1, -1)))))

@Martijn Pieters 2013-10-24 15:29:51

@dequestarmappartialsetattr: In Python 3.3 and up, the set is stored as a constant, bypassing the creation time altogether, eliminating the creation time. Tuples can be cheap to create as Python caches a bundle of them to avoid memory churn, making that the biggest difference with sets here.

@Martijn Pieters 2013-10-24 15:37:11

@dequestarmappartialsetattr: If you time just the membership test, for integers sets and tuples are equally fast for the ideal scenario; matching the first element. After that tuples lose out to sets.

@franciscod 2015-07-13 17:05:35

wow, the literal set notation is a new shiny thing for me but it's available since py2.7, on July 2010, 5 years ago! docs.python.org/3/whatsnew/2.7.html

@ShadowRanger 2016-09-04 00:37:09

@MartijnPieters: Using the set literal notation for this test isn't a savings unless the contents of the set literal are also literals, right? if 1 in {x, y, z}: can't cache the set, because x, y and z could change, so either solution needs to build a tuple or set from scratch, and I suspect whatever lookup savings you might get when checking for membership would be swamped by greater set creation time.

@Martijn Pieters 2016-09-04 07:58:07

@ShadowRanger: yes, peephole optimisation (be it for in [...] or in {...}) only works if the contents of the list or set are immutable literals too.

@Anthony 2018-07-08 21:30:02

out of curiosity, i have a list of URLs as strings whereby i'm checking if the string contains multiple substrings (i.e. 'word1' in string and 'word2' in string...). would the same syntax (i.e. string in (word1, word2):) work in this scenario or, as i am expecting, would it fail?

@Martijn Pieters 2018-07-09 11:27:58

@Anthony: that's the wrong test; string is not contained in the sequence (word1, word2). You can use all(w in string for w in sequence_of_words_to_test) if you have multiple words to test in one string. Or use the word1 in string and word2 in string approach if you have just 2 or 3 such strings to test, and they are hardcoded.

@Anthony 2018-07-09 22:35:11

okay, that's what i thought

@Dukeling 2018-09-24 12:56:13

in is not constant-time "whatever the left-hand operand is" - technically with set you'd require at most the time taken to return the hash (calculate it if you didn't cache it) and do an equals check, which can very well take longer than constant time. Although this cost also exists with tuples.

@Martijn Pieters 2018-09-24 13:12:49

@Dukeling: 'constant time' is not a fixed quantity. It means the algorithmic complexity is O(1); it doesn't matter how many elements are in the set, containment testing will take the same amount of time, roughly, independently of the size of the set.

@Martijn Pieters 2018-09-24 13:14:33

@Dukeling: containment testing against a list or tuple or other sequence takes linear time, O(N), as each element is tested one by one. So when the tuple or list or sequence doubles, so does your average time taken. Sets and dictionaries thus beat sequences, especially as the number of elements grows.

@Martijn Pieters 2018-09-24 13:15:31

@Dukeling: calculating the hash is a separate O(something) operation, which is independent of the N size of the set, dict, tuple, etc. on the right-hand side. Python immutable types cache the value anyway.

@Dukeling 2018-09-24 13:18:12

@MartijnPieters I know, but I'm still not sure why you say "whatever the left-hand operand is", because the cost changes depending on which elements you have in the set (even if it's always pseudo-constant) - I'd be fine with that paragraph if you remove that part, or replace it with "regardless of how many elements you have in the set*".

@Martijn Pieters 2018-09-24 13:50:40

@Dukeling: I really don't see a need. Bad hash implementations are the exception, not the norm. We'd have to qualify listobj[index] as O(1) (provided index is not a custom Python object with a bad __index__ implementation), or sorting a list as *O(N log N) (provided the contents do not have complex ordering implementations that throw off comparison costs). For contains operations, the cost can be assumed to be constant. If the hash cost for a 'bad actor' object is large, it'll still be the same cost independent of the size of the set you test against.

@coldspeed 2018-12-27 09:19:02

@SingleNegationElimination sure, tuples are faster than sets when comparing 2-3 values, but I would rank an answer imparting good practices from a time complexity perspective much higher than an answer that takes advantage of implementation details (which are subject to change) to achieve microptimizations on the level of "seriously, don't worry about it".

@dansalmo 2013-07-11 21:56:54

Your problem is more easily addressed with a dictionary structure like:

x = 0
y = 1
z = 3
d = {0: 'c', 1:'d', 2:'e', 3:'f'}
mylist = [d[k] for k in [x, y, z]]

@aragaer 2013-10-24 15:39:35

Or even d = "cdef" which leads to MyList = ["cdef"[k] for k in [x, y, z]]

@dansalmo 2014-05-08 14:36:49

or map(lambda i: 'cdef'[i], [x, y, z])

@dansalmo 2018-07-24 21:05:58

@MJM the output order is not determined by the dict, it is determined by the order of the list [x, y, z]

@Vinayak Kaniyarakkal 2017-06-05 06:50:28

One line solution:

mylist = [{0: 'c', 1: 'd', 2: 'e', 3: 'f'}[i] for i in [0, 1, 2, 3] if i in (x, y, z)]

Or:

mylist = ['cdef'[i] for i in range(4) if i in (x, y, z)]

@ThatGuyRussell 2015-08-19 02:34:58

Previous Solution: As stated by Martijn Pieters, the correct, and fastest, format is:

if 1 in {x, y, z}:

The one major issue that does not seem to be addressed is that you want your output list to include each letter after a true if statement.

Using only Martijn Pieters' advice you would now have:

if 0 in {x, y, z}:
    Mylist.append("c")
elif 1 in {x, y, z}:
    Mylist.append("d")
...

Problem: The first if statement would return true, and you would never get to the following elif statement. So your list would simply return:

["c"]

What you want is to have separate if statements so that python will read each statement whether the former were true or false. Such as:

if 0 in {x, y, z}:
    Mylist.append("c")
if 1 in {x, y, z}:
    Mylist.append("d")
if 2 in {x, y, z}:
    Mylist.append("e")
...

This will work, but 'if' you are comfortable using dictionaries (see what I did there), you can clean this up by making an initial dictionary mapping the numbers to the letters you want, then just using a 'for' loop:

numToLetters = {0:"c", 1:"d", 2:"e", 3:"f"}
for number in numToLetters:
    if number in {x, y, z}:
        Mylist.append(numToLetters[number])

@ytpillai 2015-05-25 03:53:36

If you ARE very very lazy, you can put the values inside an array. Such as

list = []
list.append(x)
list.append(y)
list.append(z)
nums = [add numbers here]
letters = [add corresponding letters here]
for index in range(len(nums)):
    for obj in list:
        if obj == num[index]:
            MyList.append(letters[index])
            break

You can also put the numbers and letters in a dictionary and do it, but this is probably a LOT more complicated than simply if statements. That's what you get for trying to be extra lazy :)

One more thing, your

if x or y or z == 0:

will compile, but not in the way you want it to. When you simply put a variable in an if statement (example)

if b

the program will check if the variable is not null. Another way to write the above statement (which makes more sense) is

if bool(b)

Bool is an inbuilt function in python which basically does the command of verifying a boolean statement (If you don't know what that is, it is what you are trying to make in your if statement right now :))

Another lazy way I found is :

if any([x==0, y==0, z==0])

@michael zxc858 2015-06-29 07:03:58

This code may be helpful

L ={x, y, z}
T= ((0,"c"),(1,"d"),(2,"e"),(3,"f"),)
List2=[]
for t in T :
if t[0] in L :
    List2.append(t[1])
    break;

@Saksham Varma 2015-02-27 01:31:00

d = {0:'c', 1:'d', 2:'e', 3: 'f'}
x, y, z = (0, 1, 3)
print [v for (k,v) in d.items() if x==k or y==k or z==k]

@Bhargav Boda 2015-01-13 12:10:25

I think this will handle it better:

my_dict = {0: "c", 1: "d", 2: "e", 3: "f"}

def validate(x, y, z):
    for ele in [x, y, z]:
        if ele in my_dict.keys():
            return my_dict[ele]

Output:

print validate(0, 8, 9)
c
print validate(9, 8, 9)
None
print validate(9, 8, 2)
e

@GuiltyDolphin 2014-06-04 17:09:22

To check if a value is contained within a set of variables you can use the inbuilt modules itertools and operator.

For example:

Imports:

from itertools import repeat
from operator import contains

Declare variables:

x = 0
y = 1
z = 3

Create mapping of values (in the order you want to check):

check_values = (0, 1, 3)

Use itertools to allow repetition of the variables:

check_vars = repeat((x, y, z))

Finally, use the map function to create an iterator:

checker = map(contains, check_vars, check_values)

Then, when checking for the values (in the original order), use next():

if next(checker)  # Checks for 0
    # Do something
    pass
elif next(checker)  # Checks for 1
    # Do something
    pass

etc...

This has an advantage over the lambda x: x in (variables) because operator is an inbuilt module and is faster and more efficient than using lambda which has to create a custom in-place function.

Another option for checking if there is a non-zero (or False) value in a list:

not (x and y and z)

Equivalent:

not all((x, y, z))

@wallacer 2014-06-04 17:39:09

This doesn't answer the OP's question. It only covers the first case in the provided example.

@akaRem 2013-07-11 21:16:33

The direct way to write x or y or z == 0 is

if any(map((lambda value: value == 0), (x,y,z))):
    pass # write your logic.

But I dont think, you like it. :) And this way is ugly.

The other way (a better) is:

0 in (x, y, z)

BTW lots of ifs could be written as something like this

my_cases = {
    0: Mylist.append("c"),
    1: Mylist.append("d")
    # ..
}

for key in my_cases:
    if key in (x,y,z):
        my_cases[key]()
        break

@SethMMorton 2014-02-08 20:57:33

In your example of the dict instead of a key, you will get errors because the return value of .append is None, and calling None gives an AttributeError. In general I agree with this method, though.

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