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"]
```

## Solution 1

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 (i.e. `in`

takes a fixed amount of time whatever the left-hand operand is).

### Explanation

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}`

.

## Solution 2

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]]
```

## Solution 3

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

```
if 1 in {x, y, z}:
```

Using his advice you would now 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:

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

## Solution 4

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 `if`

s 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
```

## Solution 5

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])
```

## Solution 6

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))
```

## Solution 7

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).

## Solution 8

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
```

## Solution 9

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)
```

## Solution 10

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.

## Solution 11

```
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]
```

## Solution 12

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;
```

## Solution 13

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']
```

## Solution 14

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)]
```

## Solution 15

Maybe you need direct formula for output bits set.

```
x=0 or y=0 or z=0 is equivalent to x*y*z = 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
```

Let's map to bits: `'c':1 'd':0xb10 'e':0xb100 'f':0xb1000`

Relation of isc (is 'c'):

```
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 the sum of squares of equations - logic
`or`

is the product of equations

and you'll have a 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 elements would correspond to ready string.

`array[sum]`

gives you the string.

## Solution 16

The most pythonic 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")
```

## Solution 17

It can be done easily as

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

## Solution 18

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}:`

## Solution 19

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])
```

## Solution 20

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
```

## Solution 21

Without dict, try this solution:

```
x, y, z = 0, 1, 3
offset = ord('c')
[chr(i + offset) for i in (x,y,z)]
```

and gives:

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

## Solution 22

This will help you.

```
def test_fun(val):
x = 0
y = 1
z = 2
myList = []
if val in (x, y, z) and val == 0:
myList.append("C")
if val in (x, y, z) and val == 1:
myList.append("D")
if val in (x, y, z) and val == 2:
myList.append("E")
test_fun(2);
```

## Solution 23

You can unite this

```
x = 0
y = 1
z = 3
```

in one variable.

```
In [1]: xyz = (0,1,3,)
In [2]: mylist = []
```

Change our conditions as:

```
In [3]: if 0 in xyz:
...: mylist.append("c")
...: if 1 in xyz:
...: mylist.append("d")
...: if 2 in xyz:
...: mylist.append("e")
...: if 3 in xyz:
...: mylist.append("f")
```

Output:

```
In [21]: mylist
Out[21]: ['c', 'd', 'f']
```

## Solution 24

you can develop it through two ways

```
def compareVariables(x,y,z):
mylist = []
if x==0 or y==0 or z==0:
mylist.append('c')
if x==1 or y==1 or z==1:
mylist.append('d')
if x==2 or y==2 or z==2:
mylist.append('e')
if x==3 or y==3 or z==3:
mylist.append('f')
else:
print("wrong input value!")
print('first:',mylist)
compareVariables(1, 3, 2)
```

Or

```
def compareVariables(x,y,z):
mylist = []
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')
if 3 in (x,y,z):
mylist.append('f')
else:
print("wrong input value!")
print('second:',mylist)
compareVariables(1, 3, 2)
```

## Solution 25

The `or`

does not work like that, as explained by this answer.

While the generic answer would be use

```
if 0 in (x, y, z):
...
```

this is not the best one for the *specific* problem. In your case you're doing *repeated tests*, therefore it is worthwhile to compose a *set* of these variables:

```
values = {x, y, z}
if 0 in values:
mylist.append("c")
if 1 in values:
mylist.append("d")
```

We can simplify this using a dictionary - this will result in the same values:

```
mappings = {0: "c", 1: "d", ...}
for k in mappings:
if k in values:
mylist.append(mappings[k])
```

Or if the ordering of the `mylist`

is arbitrary, you can loop over the *values* instead and match them to the mappings:

```
mappings = {0: "c", 1: "d", ...}
for v in (x, y, z):
if v in mappings:
mylist.append(mappings[v])
```

## Solution 26

### Problem

While the pattern for testing multiple values

```
>>> 2 in {1, 2, 3}
True
>>> 5 in {1, 2, 3}
False
```

is very readable and is working in many situation, there is one pitfall:

```
>>> 0 in {True, False}
True
```

But we want to have

```
>>> (0 is True) or (0 is False)
False
```

### Solution

One generalization of the previous expression is based on the answer from ytpillai:

```
>>> any([0 is True, 0 is False])
False
```

which can be written as

```
>>> any(0 is item for item in (True, False))
False
```

While this expression returns the right result it is not as readable as the first expression :-(

## Solution 27

Here is one more way to do it:

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

It is a mix of **list comprehension** and **any** keyword.

## Solution 28

**usage without if example:**

```
x,y,z = 0,1,3
values = {0:"c",1:"d",2:"e",3:"f"} # => as if usage
my_list = [values[i] for i in (x,y,z)]
print(my_list)
```

## Solution 29

FIRST, A CORRECTION TO THE `OR`

CONDITIONAL:

You need to say:

```
if x == 0 or y == 0 or z == 0:
```

The reason is that "or" splits up the condition into separate logical parts. The way your original statement was written, those parts were:

```
x
y
z == 0 // or 1, 2, 3 depending on the if statement
```

The last part was fine --- checking to see if z == 0, for instance --- but the first two parts just said essentially `if x`

and `if y`

. Since integers always evaluate to `True`

unless they're 0, that means the first part of your condition was always `True`

when `x`

or `y`

didn't equal 0 (which in the case of y was always, since you had `y = 1`

, causing your whole condition (because of how `OR`

works) to always be `True`

.

To avoid that, you need to make sure all parts of your condition (each side of the `OR`

) make sense on their own (you can do that by pretending that the other side(s) of the `OR`

statement doesn't exist). That's how you can confirm whether or not your `OR`

condition is correctly defined.

You would write the statements individually like so:

```
if x == 0
if y == 0
if z == 0
```

which means the correct mergin with the `OR`

keyword would be:

```
if x == 0 or y == 0 or z == 0
```

SECOND, HOW TO SOLVE THE PROBLEM:

You're basically wanting to check to see if any of the variables match a given integer and if so, assign it a letter that matches it in a one-to-one mapping. You want to do that for a certain list of integers so that the output is a list of letters. You'd do that like this:

```
def func(x, y, z):
result = []
for integer, letter in zip([0, 1, 2, 3], ['c', 'd', 'e', 'f']):
if x == integer or y == integer or z == integer:
result.append(letter)
return result
```

Similarly, you could use LIST COMPREHENSION to achieve the same result faster:

```
def func(x, y, z):
return [
letter
for integer, letter in zip([0, 1, 2, 3], ['c', 'd', 'e', 'f'])
if x == integer or y == integer or z == integer
]
```

## Solution 30

```
#selection
: a=np.array([0,1,3])
#list of options[sel,sel]
: np.diag(['c','d','e','f'])[a,a]
array(['c', 'd', 'f'], dtype='<U1')
```