Lab 9 Part 1

Question

My partner and I are a little lost on how to start. We can't figure out how to print out the number of complaints associated with the company name. Any suggestions?

Answer 1

I'd recommend using a for loop to pull the company name from each complaint, append it to a list, and then use the .count() method for each company in the list.

Comments

Yea I think we realized we needed a loop but we are not sure how to pull the company name from each complaint. I don't get the part where the directions say, "the list of complaints where each complaint is a dictionary."
not sure how to go about it.

---

So in the list of complaints, each complaint is a dictionary. That means listOfComplaints[0] is a dictionary object. So if you set complaint = listOfComplaints[0], you can now more easily access that dictionary. If you set company = complaint['Company'] that should give you the company name. If you look at the csv file, you'll notice each row is a dictionary, with each column's label being a key.

---

Ok, I think I get it. So those values on the very top of the text file are the keys? I am still lost on how all the rows are considered dictionaries if some don't provide a date and some other values though.

---

I think that is just considered "missing data." Dr. Stonedahl mentions that on the assignment sheet. It's still a dictionary, but with those keys having no value. I'm not 100% sure whether the value is None or the empty string or what, but fortunately, it doesn't break the whole dictionary.

---

Ah ok, thanks I was confused about that it looked like it was a continuation from another dict. Thanks.

Answer 2

First of all, you have to create a variable to count the number of complaints.

The approach is to go through each Dict of the List by using the for loop. Each Dict has the <key> labeled as "Company", while the <value> of that <key> is the actual name of the company.

Like in the PDF:

complaintDict: = {...
                  ...
                  'Company': '<name of the company>'
                  ...}

Therefore, during the loop, you want to check if the company name the user inputted matches with the <value> of the 'Company' key of each Dict. It will be easier if you set another variable to store that <value> in order to compare it with the user's input.

Every time we go through each Dict and there is a match, then the count variable will add 1 to itself.

In the end, if that variable equal to 0, the company is definitely not on that list. Otherwise, you can print it.

Comments

Would I have to make the for loop go through each dictionary then in order to find the company name for each complaint?

---

Yes, that's the only loop that you need, which is to go through each Dict inside the List