Transform dict of lists to CSV

Question

I want to transform this dict to CSV:

{
  "California": ["San Fransisco", "Los Angeles","Oakland"],
  "Texas": ["Dallas", "Houston", "Austin"],
  "Florida": ["Miami", "Tampa"],
  ...
}

I want the following output:

California,San Fransisco
California,Los Angeles
California,Oakland
Texas,Dallas
Texas,Houston
Texas,Austin
Florida,Miami
Florida,Tampa

I wrote this code. This works well, but I wonder if there is a more pythonic way to do the same.

import csv

d = {
    "California": ["San Fransisco", "Los Angeles","Oakland"],
    "Texas": ["Dallas", "Houston", "Austin"],
    "Florida": ["Miami", "Tampa"]
}

with open("./out.csv", "w") as f:
  header = ["state", "city"]
  writer = csv.writer(f)
  writer.writerow(header)
  for i in d.keys():
    for j in d[i]:
      writer.writerow([i,j])

Btw, this sort of operation is maybe called going from jagged array (I would say wide data, but the rows aren't all the same length) to tall data. — qwr
– qwr, Commented Sep 9, 2022 at 5:30

200_success · Accepted Answer · 2022-09-08 13:01:41Z

9

As per PEP 8, the standard indentation for Python code is 4 spaces. Since whitespace is significant in Python, this is a pretty strong convention that you should follow.

The code itself isn't bad, but you could make it a bit more elegant using itertools.product() and csvwriter.writerows(). You can also use more meaningful variable names than i and j.

import itertools

with open("./out.csv", "w") as f:
    writer = csv.writer(f)
    writer.writerow(["state", "city"])
    for state, cities in d.items():
        writer.writerows(itertools.product([state], cities))

answered Sep 8, 2022 at 13:01

200_success

146k22 gold badges191 silver badges481 bronze badges

4

\$\begingroup\$ I also like that you've switched to d.items() instead of d.keys(). \$\endgroup\$

Teepeemm
– Teepeemm

2022-09-08 14:22:02 +00:00
Commented Sep 8, 2022 at 14:22
1

\$\begingroup\$ That would be a more useful comment if you commented on the difference and why it's an improvement, @Teepeemm. \$\endgroup\$

Cody Gray
– Cody Gray

2022-09-09 05:16:08 +00:00
Commented Sep 9, 2022 at 5:16
1

\$\begingroup\$ Using .item() is clearly better and more readable too, if you look at the original implementation (for j in d[i]:..). Plus, applying itertools instead of a loop is smart. \$\endgroup\$

Kate
– Kate

2022-09-09 13:02:36 +00:00
Commented Sep 9, 2022 at 13:02

Add a comment |

Mateen Ulhaq · Accepted Answer · 2022-09-09 00:56:46Z

2

Flatten the dictionary into rows using clear variable names:

header = ["state", "city"]

rows = (
    [state, city]
    for state, cities in d.items()
    for city in cities
)

Then write those rows to a csv:

with open("./out.csv", "w") as f:
    writer = csv.writer(f)
    writer.writerow(header)
    writer.writerows(rows)

answered Sep 9, 2022 at 0:56

Mateen Ulhaq

1715 bronze badges

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