iterate over several collections in parallel - python

I am trying to create a list of objects (from a class defined earlier) through a loop. The structure looks something like:
ticker_symbols = ["AZN", "AAPL", "YHOO"]
stock_list = []
for i in ticker_symbols:
stock = Share(i)
pe = stock.get_price_earnings_ratio()
ps = stock.get_price_sales()
stock_object = Company(pe, ps)
stock_list.append(stock_object)
I would however want to add one more attribute to the Company-objects (stock_object) through the loop. The attribute would be a value from another list, like (arbitrary numbers) [5, 10, 20] where the first attribute would go to the first object, the second to the second object etc.Is it possible to do something like:
for i, j in ticker_symbols, list2:
#dostuff
? Could not get this sort of nested loop to work on my own. Thankful for any help.

I believe that all you have to do is change the for loop.
Instead of "for i in ticker_symbols:" you should loop like
"for i in range(len(ticker_symbols))" and then use the index i to do whatever you want with the second list.
ticker_symbols = ["AZN", "AAPL", "YHOO"]
stock_list = []
for i in range(len(ticker_symbols):
stock = Share(ticker_symbols[i])
pe = stock.get_price_earnings_ratio()
ps = stock.get_price_sales()
# And then you can write
px = whatever list2[i]
stock_object = Company(pe, ps, px)
stock_list.append(stock_object)
Some people say that using index to iterate is not good practice, but I don't think so specially if the code works.

Try:
for i, j in zip(ticker_symbols, list2):
Or
for (k, i) in enumerate(ticker_symbols):
j = list2[k]
Equivalently:
for index in range(len(ticker_symbols)):
i = ticker_symbols[index]
j = list2[index]

Related

Python: Ignore repetition in a loop

Let's say I have a function like this, to merge names in two lists:
def merge_name(list1="John",list2="Doe"):
def merge(name1=list1,name2=list2):
merge=name1+"-"+name2
data={"Merged":merge}
return data
d = pd.DataFrame()
for i,j in [(i,j) for i in list1 for j in list2]:
if i==j:
d=d
else:
x = merge(name1=i,name2=j)
ans=pd.DataFrame({"Merged":[x["Merged"]]})
d=pd.concat([d,ans])
return d
What I am interested in are unique combinations, i.e, "John-Doe" and "Doe-John" are the same for my purposes. So if I run something like this:
names1=["John","Doe","Richard"]
names2=["John","Doe","Richard","Joana"]
df=merge_name(list1=names1,list2=names2)
I will get:
John-Doe
John-Richard
John-Joana
Doe-John
Doe-Richard
Doe-Joana
Richard-John
Richard-Doe
Richard-Joana
The groups in bold are all repetitions. Essentially, every time it comes to the next i, it creates n-1 repeated groups, with n being the position in names1. Is there a way to avoid this, like drop the top name in "list2" every time j becomes the last element in the list?
Thanks in advance.
I have tried to update list2 while in loop but obviously that does not work
Below code can be useful
import pandas as pd
def merge_name(list1="John", list2="Doe"):
merged=[]
for i in list1:
for j in list2:
if (i!=j) and (f"{j} - {i}" not in merged):
merged.append(f"{i} - {j}")
df = pd.DataFrame(set(merged))
return df
names1 = ["John", "Doe", "Richard"]
names2 = ["John", "Doe", "Richard", "Joana"]
df = merge_name(list1=names1, list2=names2)
print(df)
Below is my solution with some explanations:
def combineName(listName):
res = []
for i in range(len(listName)):
for j in range(i+1, len(listName)):
res.append(listName[i] + "-" + listName[j])
return res
names1=["John","Doe","Richard"]
names2=["John","Doe","Richard","Joana"]
listName = list(set(names1 + names2))
print(listName)
print(combineName(listName))
First, you should create a simple list without repetitions. This way you only get unique elements in your list. To do this, I used a set.
I take care to transform my set into a list because later I go through the structure in a given order, which is not supposed to be true for a set.
Secondly, the function creates all the combinations. There are two loops, and you notice that the second loop has a special range.
Indeed, you do not want repetitions such as "John-Doe" and "Doe-John".
Each combination is created at a unique time!

How do I use a while loop to access all the 2nd elements of lists which are the values stored in a dictionary?

If I have a dictionary like this, filled with similar lists, how can I apply a while loo tp extract a list that prints that second element:
racoona_valence={}
racoona_valence={"rs13283416": ["7:87345874365-839479328749+","BOBB7"],\}
I need to print the part that says "BOBB7" for 2nd element of the lists in a larger dictionary. There are ten key-value pairs in it, so I am starting it like so, but unsure what to do because all the examples I can find don't relate to my problem:
n=10
gene_list = []
while n>0:
Any help greatly appreciated.
Well, there's a bunch of ways to do it depending on how well-structured your data is.
racoona_valence={"rs13283416": ["7:87345874365-839479328749+","BOBB7"], "rs13283414": ["7:87345874365-839479328749+","BOBB4"]}
output = []
for key in racoona_valence.keys():
output.append(racoona_valence[key][1])
print(output)
other_output = []
for key, value in racoona_valence.items():
other_output.append(value[1])
print(other_output)
list_comprehension = [value[1] for value in racoona_valence.values()]
print(list_comprehension)
n = len(racoona_valence.values())-1
counter = 0
gene_list = []
while counter<=n:
gene_list.append(list(racoona_valence.values())[n][1])
counter += 1
print(gene_list)
Here is a list comprehension that does what you want:
second_element = [x[1] for x in racoona_valence.values()]
Here is a for loop that does what you want:
second_element = []
for value in racoona_valence.values():
second_element.append(value[1])
Here is a while loop that does what you want:
# don't use a while loop to loop over iterables, it's a bad idea
i = 0
second_element = []
dict_values = list(racoona_valence.values())
while i < len(dict_values):
second_element.append(dict_values[i][1])
i += 1
Regardless of which approach you use, you can see the results by doing the following:
for item in second_element:
print(item)
For the example that you gave, this is the output:
BOBB7

How to take only last value from a list with unique tag?

In my LIST(not dictionary) I have these strings:
"K:60",
"M:37",
"M_4:47",
"M_5:89",
"M_6:91",
"N:15",
"O:24",
"P:50",
"Q:50",
"Q_7:89"
in output I need to have
"K:60",
"M_6:91",
"N:15",
"O:24",
"P:50",
"Q_7:89"
What is the possible decision?
Or even maybe, how to take tag with the maximum among strings with the same tag.
Use re.split and list comprehension as shown below. Use the fact that when the dictionary dct is created, only the last value is kept for each repeated key.
import re
lst = [
"K:60",
"M:37",
"M_4:47",
"M_5:89",
"M_6:91",
"N:15",
"O:24",
"P:50",
"Q:50",
"Q_7:89"
]
dct = dict([ (re.split(r'[:_]', s)[0], s) for s in lst])
lst_uniq = list(dct.values())
print(lst_uniq)
# ['K:60', 'M_6:91', 'N:15', 'O:24', 'P:50', 'Q_7:89']
Probably far from the cleanest but here is a method quite easy to understand.
l = ["K:60", "M:37", "M_4:47", "M_5:89", "M_6:91", "N:15", "O:24", "P:50", "Q:50", "Q_7:89"]
reponse = []
val = []
complete_val = []
for x in l:
if x[0] not in reponse:
reponse.append(x[0])
complete_val.append(x.split(':')[0])
val.append(int(x.split(':')[1]))
elif int(x.split(':')[1]) > val[reponse.index(x[0])]:
val[reponse.index(x[0])] = int(x.split(':')[1])
for x in range(len(complete_val)):
print(str(complete_val[x]) + ":" + str(val[x]))
K:60
M:91
N:15
O:24
P:50
Q:89
I do not see any straight-forward technique. Other than iterating on entire thing and computing yourself, I do not see if any built-in can be used. I have written this where you do not require your values to be sorted in your input.
But I like the answer posted by Timur Shtatland, you can make us of that if your values are already sorted in input.
intermediate = {}
for item in a:
key, val = item.split(':')
key = key.split('_')[0]
val = int(val)
if intermediate.get(key, (float('-inf'), None))[0] < val:
intermediate[key] = (val, item)
ans = [x[1] for x in intermediate.values()]
print(ans)
which gives:
['K:60', 'M_6:91', 'N:15', 'O:24', 'P:50', 'Q_7:89']

Python: Concatenate similiar objects in List

I have a list containing strings as ['Country-Points'].
For example:
lst = ['Albania-10', 'Albania-5', 'Andorra-0', 'Andorra-4', 'Andorra-8', ...other countries...]
I want to calculate the average for each country without creating a new list. So the output would be (in the case above):
lst = ['Albania-7.5', 'Andorra-4.25', ...other countries...]
Would realy appreciate if anyone can help me with this.
EDIT:
this is what I've got so far. So, "data" is actually a dictionary, where the keys are countries and the values are list of other countries points' to this country (the one as Key). Again, I'm new at Python so I don't realy know all the built-in functions.
for key in self.data:
lst = []
index = 0
score = 0
cnt = 0
s = str(self.data[key][0]).split("-")[0]
for i in range(len(self.data[key])):
if s in self.data[key][i]:
a = str(self.data[key][i]).split("-")
score += int(float(a[1]))
cnt+=1
index+=1
if i+1 != len(self.data[key]) and not s in self.data[key][i+1]:
lst.append(s + "-" + str(float(score/cnt)))
s = str(self.data[key][index]).split("-")[0]
score = 0
self.data[key] = lst
itertools.groupby with a suitable key function can help:
import itertools
def get_country_name(item):
return item.split('-', 1)[0]
def get_country_value(item):
return float(item.split('-', 1)[1])
def country_avg_grouper(lst) :
for ctry, group in itertools.groupby(lst, key=get_country_name):
values = list(get_country_value(c) for c in group)
avg = sum(values)/len(values)
yield '{country}-{avg}'.format(country=ctry, avg=avg)
lst[:] = country_avg_grouper(lst)
The key here is that I wrote a function to do the change out of place and then I can easily make the substitution happen in place by using slice assignment.
I would probabkly do this with an intermediate dictionary.
def country(s):
return s.split('-')[0]
def value(s):
return float(s.split('-')[1])
def country_average(lst):
country_map = {}|
for point in lst:
c = country(pair)
v = value(pair)
old = country_map.get(c, (0, 0))
country_map[c] = (old[0]+v, old[1]+1)
return ['%s-%f' % (country, sum/count)
for (country, (sum, count)) in country_map.items()]
It tries hard to only traverse the original list only once, at the expense of quite a few tuple allocations.

looping dictionaries of {tuple:NumPy.array}

i have a set of dictionaries k of the form {(i,j):NumPy.array} over which I want to loop the NumPy.arrays for a certain evaluation.
I made the dictionarries as follows:
datarr = ['PowUse', 'PowHea', 'PowSol', 'Top']
for i in range(len(dat)): exec(datarr[i]+'={}')
so i can always change the set of data i want to evaluate in my bigger set of code by changeing the original list of strings. However, this means i have to call for my dictionaries as eval(k) for k in datarr.
As a result, the loop i want to do looks like this for the moment :
for i in filarr:
for j in buiarr:
for l in datarrdif:
a = eval(l)[(i, j)]
a[abs(a)<.01] = float('NaN')
eval(l).update({(i, j):a})
but is there a much nicer way to write this ? I tried following, but this didn't work:
[eval(l)[(i, j)][abs(eval(l)[(i, j)])<.01 for i in filarr for j in buiarr for k in datarrdiff] = float('NaN')`
Thx in advance
datarr = ['PowUse', 'PowHea', 'PowSol', 'Top']
for i in range(len(dat)): exec(datarr[i]+'={}')
Why don't you create them as a dictionary of dictionaries?
datarr = ['PowUse', 'PowHea', 'PowSol', 'Top']
data = dict((name, {}) for name in datarr)
Then you can avoid all the eval().
for i in filarr:
for j in buiarr:
for l in datarr:
a = data[l][(i, j)]
np.putmask(a, np.abs(a)<.01, np.nan)
data[l].update({(i, j):a})
or probably just:
for arr in data.itervalues():
np.putmask(arr, np.abs(arr)<.01, np.nan)
if you want to set all elements of all dictionary values where abs(element) < .01 to NaN .

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