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How do I convert def stringParametre(x) x to a list so that if x is hello then it will become a list like ["H","E","l","l","O"] . Capitals don't matter .
Note that you do not have to convert to a list if all you want to do is to iterate over the characters of the string:
for c in "hello":
# do something with c
works
Building on #idjaw's comment:
def stringParametre(x):
return list(x)
Of course this will have an error if x is not a string (or other sequence type).
list(x)
OR
mylist = []
for c in x:
mylist.append(c)
Related
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I have an output of a = "[1,2,3]", how do I convert it to an array a = [1,2,3] from the string in Python?
Thanks!
Try this:
import json
result = json.loads("[1,2,3]")
print(result)
Use ast and then ast.literal_eval()
import ast
print(ast.literal_eval(x))
You can do it without importing stuff
lst = []
for index in range(len(a)):
if a[index].isdigit():
lst.append(int(a[index]))
Then lst will be a list containing all the integers in string a. And if you want to keep the variable name "a", you can
a = lst [::]
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I have a list of strings L.
I need to check, whether a string is either directly an element of L or is in this format: "foo-element_of_L"
Is there a better way to do this in python than adding "foo-X" to L for all X in L?
I would do two lookups:
if x in L or f'foo-{x}' in L:
which may be significantly faster than
if any(x == y or f'foo-{x}' == y for x in L):
which is essentially what you were proposing.
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I have a list in python of the following form:
myList = ['r0x94', 'r0x21', 'r0x51']
I want to sort it based on the last number in each string entry of the list such that:
sorted_myList = ['r0x21', 'r0x51', 'r0x94']
The last number is not hex, rather it is decimal. How to do it?
>>> my_list = ['r0x94', 'r0x21', 'r0x51']
>>> sorted(my_list, key=lambda x: int(x.rpartition('x')[-1]))
['r0x21', 'r0x51', 'r0x94']
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For example, I have this:
alphabetValues = {"a":1,"b":2,"c":3,"d":4,"e":5,"f":6,"g":7...
Is it possible if instead of having:
print(alphabetValues["c"])
To having something that would get "e" if I searched for 5 in a dict.
"e":5
Thanks in advance.
As suggested by jonrsharpe, you need to reverse your dictionnary :
alphabetValues = {"a":1,"b":2,"c":3,"d":4,"e":5,"f":6,"g":7}
revalpha={v:k for k,v in alphabetValues.iteritems()}
>>> revalpha[5]
'e'
Why not set up an alphabet list?
alphabet = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
print(alphabet[0]) #will print out 'a'
print(alphabet[25]) #will print out 'z'
Note that all values are 1 less than expected.
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I am making a code breaker game on python and need to make a if statement that checks whether the list contains any symbols or numbers.
So it would tell me that:
list=[hello, bonjour, 4 , hola]`
contains a number. And:
list2=[hello, bonjour, hola]
does not contain a number
Just use:
>>> test = 'abc123%def'
>>> any(not x.isalpha() for x in test)
True
>>> test2 = 'abc'
>>> any(not x.isalpha() for x in test2)
False