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How can I write following while loop in python?
int v,t;
while (scanf("%d %d",&v,&t)==2)
{
//body
}
Python being a higher level language than C will normally use different patterns for this kind of situation. There is a deleted question that in fact had the exact same behavior than your C snippet - and thus would be "correct", but it became so ugly a piece of code in Python it was downvoted pretty fast.
So, first things first - it is 2015, and people can't simply look at a C's "scanf" on the prompt and divine they should type two white space separated integer numbers - you'd better give then a message. Anther big difference is that in Python variable assignments are considered statements, and can't be done inside an expression (the while expression in this case). So you have to have a while expression that is always true, and decide whether to break later.
Thus you can go with a pattern like this.
while True:
values = input("Type your values separated by space, any other thing to exit:").split()
try:
v = int(values[0])
t = int(values[1])
except IndexError, ValueError:
break
<body>
This replaces the behavior of your code:
import re
while True:
m = re.match("(\d) (\d)", input()):
if m is None: #The input did not match
break #replace it with an error message and continue to let the user try again
u, v = [int(e) for e in m.groups()]
pass #body
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I have a simple program that takes data from the user. Here is an abbreviated version of it:
a = "0-1"
b = "0‑1"
print(a in b) # prints False
Problem:
ord('-') for a = 45
ord('‑') for b = 8209
How can I make sure that the "-" sign is always the same and checking a in b returns True?
The most robust way would be to use the unidecode module to convert all non-ASCII characters to their closest ASCII equivalent automatically.
import unidecode
print(unidecode.unidecode(a) in unidecode.unidecode(b))
It's not clear if your example is part of a more general, but for the example provided you can handle this using replace:
a = "0-1"
b = "0‑1"
print(a.replace("‑", "-") in b.replace("‑", "-")) # True
I've called replace on both sides, because it's not clear which side is your input and which is not. In principle though this comes down to "sanitize your input".
If this is more of a general problem, you might want to look at using .translate to produce a mapping of characters to apply in one go.
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I've coded a check for integers which returns a boolean value:
def NoZerosFives(n: str):
if '0' in n or '5' in n: #It checks if any of the digits is 0 or 5
return False
return True
which I then put in a script:
for n in range(10**20000, 10**100000000):
d = str(n)
if NoZerosFives(d):
pass
else:
print('Failed the check.')
n is obviously extremely large, and the else:print('Failed the check.') is here just to tell me if the script's going or not, but it's not printing anything.
My first thought is that the problem is not with Python but with my hardware (I have a pretty powerful laptop, but still a laptop). Is my computer not powerful enough? Should I have done something to my PC that I didn't do?
Or is the code not fit for the objective and I just need to do something different?
As #jordanm points out in a comment, NoZerosFives will not return True (which is the only way your code produces any output at all) for a very long time. Think about how long it's going to take, starting from 10**20000 (shown below), to reach a value with no zeros in it by just adding 1 at each step.
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I have a string containing a number, starting with x, for example, "x270" or "x9". It always starts with x. I need to get the number.
I'm doing this:
blatz = "x22"
a1 = int(re.search("x(\d+)", blatz).group(1))
This doesn't seem very Pythonic. I would welcome more elegant solutions.
Using re library seems to be an overkill. You don't have to search for a pattern, because you're saying that each string starts with x.
So you simply can do slicing:
blatz = "x22"
a1 = int(blatz[1:])
If you need further checks, you can look at str.startswith(), str.endswith and/or str.isdigit().
While slicing looks very pythonistic, there is also the possibility to use other string methods that lead to the same goal:
blatz = "x22"
a2 = int(blatz.lstrip("x")) # strip "x" from the left
a3 = int(blatz.partition("x")[-1]) # get everything after "x"
a4 = int(blatz.replace("x", "")) # replace every "x" with empty string
...
But slicing is faster and nothing unusual for Python programmers.
Please check out this.
import re
blatz = "x22"
print(re.search("\d", blatz).group())
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I'm currently scraping some tables on the internet where numbers are posted in varying numerical formats:
Animal - Left in Wild
Tigers - 18
Deer - 18m
Pigs - 180000
I've managed to strip the m away from the number, but I am wondering if/how I could use a if statement to allow some manipulation to ensure I document the number accurately:
if animal.strip("m") == animal.strip("m"):
left_in_wild = left_in_wild * 1000000
Obviously that code does not work, but it is a rough thought of how I'm thinking about getting around this. If anyone can has anything they think can be helpful let me know.
Thank you!
A simple IF statement could help with what you're looking for:
animal = "18m"
if 'm' in animal:
print animal.strip('m') + ",000,000"
if 'k' in animal:
print animal.strip('k') + ",000"
returns:
18,000,000
Something like:
import re
def get_number(s):
try:
i=int(re.match('(\d+)', s).group(1))
if "m" in s:
i*=1000000
return i
except:
print "No Number"
get_numbers("18m") returns 18000000
You could even expand it to have an elif "k" in s block if you had thousands or something.
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So, I have a really weird question that has kinda always bugged me. In this example:
patterns = [ 'this', 'that' ]
text = 'Does this text match the pattern?'
for pattern in patterns:
print('Looking for "%s" in "%s" ->' % (pattern, text))
So this is just a example, but what I'm wondering, is in the for loop, pattern was never declared, and I know that Python is a dynamic language so you dont have to declare variables, but how does python like know what it means? I've seen this alot with for loops and alot of the time it just seems like people put whatever they want in that part of the for loop, and I really don't get it. Does it matter what you put there?
With for loops, you're iterating over what is in patterns. What it will do is assign the first object in the list patterns to the variable pattern. And then do it again with the next object.
If you run that code, you get the following output:
Looking for this in Does this text match the pattern?
Looking for that in Does this text match the pattern?
^
|
The arrow points to the variable pattern, which changes every time the loop restarts. I should also point out that this is how the for loop runs in basically every language, including C, Java and Python.
Another simple example to demonstrate this would be to iterate over a python list of integers, done through the range() function:
for i in range(5): # range creates this: [0, 1, 2, 3, 4]
print(i)
Output:
0
1
2
3
4