In a dictionnary, how to check that a key has exactly 1 value?

Question

I have this dictionnary:

{128: ['S', 'S', 'O', 'F'], 512: ['S', 'F']}

I would like to be sure that each key has exactly one value 'F' and one value 'S' and return a message if it's not the case

I tried this but it didn't seem to work: it didn't print the message

for key in d:
    if not re.search(r"[F]{1}","".join(d[key])) or not re.search(r"[S].{1}","".join(d[key])):
        print(f"There is no start or end stop for the line {key}.")

Thanks

It does not look like a regex question, you are checking if a string equals some value. Also, your keys are numbers, there are no letters. — Wiktor Stribiżew, Nov 23 '21 at 12:52

riquefr · Answer 1 · 2021-11-23T14:33:05.153

2

Your dictionary contain a list, not a string so you shouldn't use regex. You can check if the list contain exactly the number of values you want using list.count(value).

for key in d:
  if not (d[ḱey].count('F') == 1 and d[ḱey].count('S') == 1):
    print(f"There is no start or end stop for the line {key}.")

edited Nov 23 '21 at 14:33

answered Nov 23 '21 at 12:57

riquefr

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'i' is not defined here! – AziMez Nov 23 '21 at 13:10
I took that part of code from who made the question – riquefr Nov 23 '21 at 13:12
The question was edited, now it's `key` instead of `i`. To avoid confusion you should make the same change. – Mark Ransom Nov 23 '21 at 14:31

score 0 · Answer 2 · answered Nov 23 '21 at 12:57

0

You can check the count of those characters in each value. One approach is below

sample = {128: ['S', 'S', 'O', 'F'], 512: ['S', 'F']}
wrong_lines = [index for index, value in sample.items() if value.count('S') == value.count('F') == 1]
print(f"Wrong lines {wrong_lines}")

answered Nov 23 '21 at 12:57

Kris

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This code return: **Wrong lines [512]**, which is not! – AziMez Nov 23 '21 at 13:08
it returns the key of the wrong list! – Kris Nov 23 '21 at 13:10

score 0 · Answer 3 · answered Nov 25 '21 at 11:30

0

d = {128: ['S', 'S', 'O', 'F'], 512: ['S', 'F']}

for key in d:
  if not (d[key].count('F') == 1 and d[key].count('S') == 1):
    print(f"There is no start or end stop for the line {key}.")

answered Nov 25 '21 at 11:30

sogand

1

This appears to be an exact duplicate of the code from [@riquefr's answer](https://stackoverflow.com/a/70081334/3025856), posted a couple days ago, except that you included less explanation. Please review existing answers before contributing a new one. It is nice that you included sample data, but the function itself appears to be an exact copy. If I'm missing something, others will too: In that case, edit your answer to explain what differentiates your approach. – Jeremy Caney Nov 26 '21 at 01:05

score -1 · Answer 4 · answered Nov 23 '21 at 12:55

You can use a Counter and get the counts of all letters simultaneously:

>>> from collections import Counter
>>> di={128: ['S', 'S', 'O', 'F'], 512: ['S', 'F']}
>>> {k:Counter(v) for k,v in di.items()}
{128: Counter({'S': 2, 'O': 1, 'F': 1}), 512: Counter({'S': 1, 'F': 1})}

In a dictionnary, how to check that a key has exactly 1 value?

4 Answers4