List comprehension instead of nested for loop

Question

I have this which works, but I would like to put this into a list comprehension to save computing power

emp_id=df_hours["EMPLOYEE"]
emp_list=df_item["EMPLOYEE"]

##############
#loop through and get values for the day for hours worked per server, and add them to building df
#############

hours_worked=df_hours["HOURS"]
emp_found=[]
 
for x in range(len(emp_list)):
    for i in range(len(emp_id)):
        if emp_list[x] == emp_id[i]:
            emp_found.append(hours_worked[i])

inex=0
last=len(emp_list)
hours_wored=emp_found[inex:last]
df_item['HRS']=hour_worked

score 0 · Answer 1 · answered Sep 26 '20 at 06:10

0

I believe this is what you're looking for

emp_found = [hours_worked[i] for i in range(len(emp_id)) for x in range(len(emp_list)) if emp_list[x] == emp_id[i]]

answered Sep 26 '20 at 06:10

Jacobjanak

307
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will try it, thanks – Clay Franklin Sep 26 '20 at 07:06
yes this does work. Does this use less resources or does it compute in the same amount of time and use less computing power? – Clay Franklin Sep 26 '20 at 07:28
Glad it worked! Both ways of coding this require the same amount of calculations so I think there will be no difference. I actually think the way you have it is more readable haha – Jacobjanak Sep 27 '20 at 21:18

score 0 · Answer 2 · answered Sep 26 '20 at 06:12

0

Here's a shorter, more elegant way:

print(list(set(df_hours["EMPLOYEE"]).intersection(set(df_item["EMPLOYEE"]))))

Since what you are looking for is getting a list with all common values, you can transform each column to a set, and get the intersection. Then you can convert back the result to a list.

I find this syntax shorter and you might say; more pythonic.

answered Sep 26 '20 at 06:12

snatchysquid

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thank you, haven't tried it yet - but still - thank you – Clay Franklin Sep 26 '20 at 07:05
this doesn't error, but it doesn't join the intersection, and I'm left with an empty list – Clay Franklin Sep 26 '20 at 07:22

List comprehension instead of nested for loop

2 Answers2