How to form a list from queried data? - python

I have the following code:
s = (f'{item["Num"]}')
my_list = []
my_list.append(s)
print(my_list)
As you can see i want this to form a list that i will then be able to store under my_list, the output from my code looks like this (this is a sample from around 2000 different values)
['01849']
['01852']
['01866']
['01883']
etc...
This is not what i had in mind, i want it to look like this
[`01849', '01852', '01866', '01883']
Has anyone got any suggestions on what i do wrong when i create the list? Thanks

You can fix your problem and represent this compactly with a list comprehension. Assuming your collection is called items, it can be represented as such, without the loop:
my_list = [f'{item["Num"]}' for item in items]

You should first initialize a list here, and then use a for-loop to populate it. So:
my_list = []
for values in range(0, #length of your list):
s = (f'{item["Num"]}')
my_list.append(s)
print(my_list)
Even better, you can also use a list comprehension for this:
my_list = [(f'{item["Num"]}') for values in range(0, #length of your list)]

Related

How to convert a list made up of one tuple into a two item list?

I have a function that returns a one item list, like so:
list = [('array_1','array_2')]
I want to change this so that the list is instead a two item one, without the parentheses or single quotes:
list = [array_1,array_2]
What would be the best way to go about doing this?
Try this
lists = [('array_1','array_2')]
print([y for x in lists for y in x])
output
['array_1', 'array_2']
Use chain.from_iterable
from itertools import chain
list(chain.from_iterable([('array_1','array_2')]))
['array_1', 'array_2']
You can try this:
lst_tuple = [('array_1', 'array_2')]
lst = []
for i in lst_tuple[0]:
lst.append(i)
By iteratating over the list that contains the tuple and appending each item to a new list, you can get this result:
['array_1', 'array_2']
You could just typecast like this.
list = list([('array_1','array_2')][0])

Quicker way to filter lists based on a check to external variable?

I have a variable = 'P13804'
I also have a list like this:
['1T9G\tA\t2.9\tP11310\t241279.81', '1T9G\tS\t2.9\tP38117\t241279.81', '1T9G\tD\t2.9\tP11310\t241279.81', '1T9G\tB\t2.9\tP11310\t241279.81', '1T9G\tR\t2.9\tP13804\t241279.81', '1T9G\tC\t2.9\tP11310\t241279.81']
You can see, if you split each item in this list up by tab, that the third item in each sub-list of this list is sometimes 'P11310' and sometimes is 'P13804'.
I want to remove the items from the list, where the third item does not match my variable of interest (i.e. in this case P13804).
I know a way to do this is:
var = 'P13804'
new_list = []
for each_item in list1:
split_each_item = each_item.split('\t')
if split_each_item[3] != var:
new_list.append(each_item)
print(new_list)
In reality, the lists are really long, and i have a lot of variables to check. So I'm wondering does someone have a faster way of doing this?
It is generally more efficient in Python to build a list with a comprehension than repeatedly appending to it. So I would use:
var = 'P13804'
new_list = [i for i in list1 if i.split('\t')[2] == var]
According to timeit, it saves more or less 20% of the elapsed time.

Relationship between elements of two list: how to exploit it in Python?

SO here is my minimal working example:
# I have a list
list1 = [1,2,3,4]
#I do some operation on the elements of the list
list2 = [2**j for j in list1]
# Then I want to have these items all shuffled around, so for instance
list2 = np.random.permutation(list2)
#Now here is my problem: I want to understand which element of the new list2 came from which element of list1. I am looking for something like this:
list1.index(something)
# Basically given an element of list2, I want to understand from where it came from, in list1. I really cant think of a simple way of doing this, but there must be an easy way!
Can you please suggest me an easy solution? This is a minimal working example,however the main point is that I have a list, I do some operation on the elements and assign these to a new list. And then the items get all shuffled around and I need to understand where they came from.
enumerate, like everyone said is the best option but there is an alternative if you know the mapping relation. You can write a function that does the opposite of the mapping relation. (eg. decodes if the original function encodes.)
Then you use decoded_list = map(decode_function,encoded_list) to get a new list. Then by cross comparing this list with the original list, you can achieve your goal.
Enumerate is better if you are certain that the same list was modified using the encode_function from within the code to get the encoded list.
However, if you are importing this new list from elsewhere, eg. from a table on a website, my approach is the way to go.
You could use a permutation list/index :
# I have a list
list1 = [1,2,3,4]
#I do some operation on the elements of the list
list2 = [2**j for j in list1]
# Then I want to have these items all shuffled around, so for instance
index_list = range(len(list2))
index_list = np.random.permutation(index_list)
list3 = [list2[i] for i in index_list]
then,with input_element:
answer = index_list[list3.index(input_element)]
Based on your code:
# I have a list
list1 = [1,2,3,4]
#I do some operation on the elements of the list
list2 = [2**j for j in list1]
# made a recode of index and value
index_list2 = list(enumerate(list2))
# Then I want to have these items all shuffled around, so for instance
index_list3 = np.random.permutation(index_list2)
idx, list3 = zip(*index_list3)
#get the index of element_input in list3, then get the value of the index in idx, that should be the answer you want.
answer = idx[list3.index(element_input)]
def index3_to_1(index):
y = list3[index]
x = np.log(y)/np.log(2) # inverse y=f(x) for your operation
return list1.index(x)
This supposes that the operations you are doing on list2 are reversible. Also, it supposes that each element in list1 is unique.

Delete a specific value from multi-value dictionary in python

``I have a dictionary in python like this.
dictionary = {"00":[1,2,3,4,5,6,7,8,9],"01":[1,2,3,4,5,6,7,8,9],"02":[1,2,3,4,5,6,7,8,9],"03":[1,2,3,4,5,6,7,8,9],"04":[1,2,3,4,5,6,7,8,9]........up-to "99":[1,2,3,4,5,6,7,8,9]}
I have to delete the value 2 from the list of "00".I tried it using following statement.
del (dictionary[key][dictionary[key].index(sudokumatrix[i][iindex])]).
Here key has value "00" and sudokumatrix[i][iindex] has value 2.But i got resulting dictionary as
{"00":[1,3,4,5,6,7,8,9],"01":[1,3,4,5,6,7,8,9],"02":[1,3,4,5,6,7,8,9],"03":[1,3,4,5,6,7,8,9],"04":[1,3,4,5,6,7,8,9].....}.
I have to get the result as:
{"00":[1,3,4,5,6,7,8,9],"01":[1,2,3,4,5,6,7,8,9],"02":[1,2,3,4,5,6,7,8,9],"03":[1,2,3,4,5,6,7,8,9],"04":[1,2,3,4,5,6,7,8,9]....}
I am posting the whole code here:
dictionary = dict()
zerotonine = "123456789"
list2 = list(zerotonine)
list2 = [int(i) for i in list2]
sudokumatrix=[]
for p in range(9):
for q in range(9):
keyis=str(p)+str(q)
dictionary[keyis] = list2
for i in range(9):
initialinput = [1,2,3,4,5,6,7,8,9]
list1=list(initialinput)
list1 = [int(i) for i in list1]
sudokumatrix.append(list1)
key = "00"
del dictionary[key][dictionary[key].index(sudokumatrix[0][1])]
print dictionary
EDIT == I guess(since the generation of dictionary is not given) ==EDIT.
The reason is that the values of keys '00', '01', ... are pointing to the same list. Modifying one of them will definitely affect the others.
Try using this to generate your dict
dictionary = dict((str(x).zfill(2), range(1, 10)) for x in range(100))
Your code of this part is actually not wrong, but to use list.remove() will make it much better.
The issue has to do with pointers.
replace this:
dictionary[keyis] = list2
with this:
dictionary[keyis] = [int(i) for i in list2]
You're creating list2 correctly, but when you go into the loop Python doesn't make a brand new copy of it with every iteration. It makes a pointer to the original list. Python sees:
dictionary[keyis] = list2
and says "oh, list2? I recognize that name! I have that as an object in memory already! I'll save some space by just updating the original copy and linking it here! Any time someone wants to view it or update it I'll just deal with the original and everything will be awesome forever!!!"
OK, so maybe the python interpreter isn't that enthusiastic, but that's how I like to think of it. The end result is that all of your dictionary values are pointing at the original list.
If you don't mind deleting every occurrence of 2 in the list, you can use list comprehension:
dictionary["00"] = [i for i in dictionary["00"] if i != 2]
This will create a new list, and will avoid altering the other values, as it appears all your dictionary values reference the same list.
EDIT: Yep your dictionary values reference the same list
you could use dictionary and list comprehension to create your dictionary
dictionary = {str(x):[i for i in range(10)] for x in range(100)}

Output loop result into a list

Someone please help me with this code:
agerlist = for agern in anime.info['Genres']:
print agern['name']
Is there any way to save the output into one variable? An example would be: alist = agern in anime.info ?
You mean like this?
alist = [ x['name'] for x in anime.info['Genre']]
Create an empty list and then append the results of for loop to that list or you could use list comprehension.
l = []
for agern in anime.info['Genres']:
l.append(agern['name'])
print l
you could also store that into a generator function, something like this:
for agern in anime.info['Genres']:
yield agern['name']
so that you will actually compute this thing only on request, i.e. during a for cycle, so you will not store in memory each agern['name'] two times.

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