I am trying to add the current dictionary to a Redis list using a dictionary comprehension and then to print out the first (aka current) keys and values of that list. I say current because this is a process I will be continuing with a while loop to have the list building over time, but I have to always access the first keys/values.
I am sure I am totally butchering this, but this is what I have:
adict = {"a":1,"b":2,"c":3}
{rserver.rpush("list",value) for value in adict}
print(float(rserver.lindex("list",0)))
I need to get a list of both keys and values back.
Help would be MUCH appreciated. Thanks!
I am not quite positive on what your redis-list should contain (please include your expected result in the question), but assuming it should at the end of inserts look something like this ["a:1", "b:1", "c:1"], you can achieve this with
adict = {"a":1,"b":2,"c":3}
for key,value in adict.items():
rserver.rpush("list", ":".join([key, value]))
print(float(rserver.lindex("list",0))) #>>> "a:1"
(as you have not included what interface rserver exactly is, it is a bit hard to guess on its exact behavior)
Related
As Zen of Python says:
There should be one– and preferably only one –obvious way to do it.
I can't decide which one of the following lines is better:
ps.cfg is a python dict
if cfg.get("runner") is None:
if "runner" not in cfg:
Is one of them obviously better, or could we say that they are both OK?
It depends on what you want to achieve. If you want to make sure that an element at a certain key exists in that dictionary, i.e. something was inserted at that key, you should utilize capabilities that tell you exactly that. Like in your case:
if "runner" not in cfg:
If you really don't care if actually an element at a certain key exists in that dictionary and you just want to see if trying to retrieve an element of the dictionary will give you None as result, you can also work with the None check.
Just consider that by calling get('some_key') and checking the returned value against None this won't tell you if there is an element existing at key some_key. Because no one prevents you from inserting an element without a value (meaning the value None) into a dictionary at that very key.
I have a simple problem. I'm new to python and programming so I think i miss something.
The variable "account_info" is assigned earlier and is a list of lists with 4 elements each. The variable current is a user input value, which (should) appear as the first element of the lists in the list account_info.
I want to iterate over the lists in the list and compare if the first element is equal to "current".
This is the code:
for i in account_info:
if current == account_info[i][0]:
email = account_info[i][1]
additional = account_info[i][2]
pw = account_info[i][3]
print(email)
I get an error in pycharm, when running that code. It seems that I can't iterate over the lists like that, can please someone explain and show a different solution?
Thank you
As #ForceBru commented, your issue is due to how for loops in Python work. The value you get from the loop is not an index into the iterable object you're looping on, rather, it's a value from the iterable. That makes your indexing with it later almost certainly wrong (though in certain contexts it might make sense, if you have a list that contains indexes into itself).
In your case, you probably want to do something more like this:
for account in accounts_info:
if current == account[0]: # note, only the inner indexing is needed
email = account[1]
additional = account[2]
pw = account[3]
Since you're expecting the inner lists to contain four values, you could even unpack the account values that you get from iterating directly into the inner variables. Though this would happen unconditionally, so it might not do what you want. Here's what that would look like, with the print call you were doing after the loop instead moved inside the conditional (so you only print the one email address that corresponds to the value in current):
for account_id, email, additional, pw in account_info: # unpack unconditionally
if account_id == current: # use the convenient name here
print(email) # print only in the conditional
In the rare case where you really do need to iterate over indexes, you can use the range type, which behaves like a sequence of integers (starting at zero by default). So you could replace your loop with this version and the body would work as you had intended (though this is less idiomatic Python than the previous versions).
for i in range(len(accounts_info)):
If you need both the index and the current value, you can use the enumerate function, which yields 2-tuples of index and value as you iterate over it. This is often handy when you need to reassign values in a list some times:
for i, account in enumerate(accounts_info):
if account[0] == current:
accounts_info[i] = new_value # replace the whole account entry
I have an assignment, similar to scrabble. I have to check if a subset is in the set. can only use a letter once. so if subset has 2t and the set has 1t it is false.
My problem is, I used 2 inputs to allow people to enter the subset and set, but that create a string no breaks between the letters which mean split or list won't create a LIST with individual letters. (at least I can't find any way.)
My plan was something like
wordset = word.lower().split()
subset = letters.lower()
for i in range(len(subset)):
if i in subset and in set:
set.remove(i)
I know that properly won't work but until I can get it into a list or someone gives me a hint how to do it with string I can't start testing it. Sorry for so much writing.
If you wish to get a list of characters in a given string you can use a list comprehension:
characters = [x for x in some_string]
listname = ['bartian', 'lenana', 'kilimanjaro', 'uhuru', 'elgon', 'everest']
while True:
if listname[:-1] == everest:
print(listname[:-1]+=)
I need a way to loop over this list and print the items in the list. I dont know whether this is pythonic since am having trouble printing the items. I want it to print from the last to first, middle item to lst or to first.
I think the easiest way to solve your task is by using a for loop on a slice of your original list.
for name in listname[::-1]:
print(name)
This prints the names in reverse order. The [::-1] slice says to go from one end to the other, with a step size of -1. For this specific case, you could also use reversed, but if you want to do other kinds of slicing it might make sense to use a slice here too, for symmetry.
Here are slices for the other forms you wanted:
for name in listname[len(listname)//2:]: # iterate on a slice from middle to end
print(name)
for name in listname[len(listname)//2::-1]: # slice from the middle back to the start
print(name)
Your code also has an if statement in it, but you don't describe what you want for it to do. If you're only printing names that are "everest" there's not much point to the loop!
I have a string like this:
val a = "some random test message"
I have a list like this:
val keys = List("hi","random","test")
Now, I want to check whether the string a contains any values from keys. How can we do this using the in built library functions of Scala ?
( I know the way of splitting a to List and then do a check with keys list and then find the solution. But I'm looking a way of solving it more simply using standard library functions.)
Something like this?
keys.exists(a.contains(_))
Or even more idiomatically
keys.exists(a.contains)
The simple case is to test substring containment (as remarked in rarry's answer), e.g.
keys.exists(a.contains(_))
You didn't say whether you actually want to find whole word matches instead. Since rarry's answer assumed you didn't, here's an alternative that assumes you do.
val a = "some random test message"
val words = a.split(" ")
val keys = Set("hi","random","test") // could be a List (see below)
words.exists(keys contains _)
Bear in mind that the list of keys is only efficient for small lists. With a list, the contains method typically scans the entire list linearly until it finds a match or reaches the end.
For larger numbers of items, a set is not only preferable, but also is a more true representation of the information. Sets are typically optimised via hashcodes etc and therefore need less linear searching - or none at all.