rlvaugh
commented
5 years ago Hi David: Thank you for the feedback! I'll have more time to look at this tomorrow but I quickly ran the program and I was able to find "dad" in "David": == RESTART: C:\Python35\Python 3 Stuff\Anagrams\for_book\phrase_anagrams.py == Enter a name: david Length of anagram phrase = 0 a ad add aid avid dad did diva i id viaRemaining letters = david Number of remaining letters = 5 Number of remaining (real word)anagrams = 11 Current anagram phrase =
If you've downloaded the book's code from https://nostarch.com/impracticalpythonprojects, could you run the phrase_anagrams.py program found there? When the publisher moved my manuscript from Word to InDesign, it messed up some of the code printed in the book, especially indents. For other projects, please check the "Update" section on the website above for the current errata. I may be misreading your note; I'm short for time right now. Let me know if this doesn't help. Regards,
Lee
On Sunday, February 17, 2019 08:55:56 PM CST, dwanneruchi <notifications@github.com> wrote: @rlvaugh this book is great! Really enjoying it!
It looks like find_anagrams() is only able to produce anagrams composed of unique letters (i.e. n letters with n distinct letters).
For example, my name is david. When I run this function the word "dad" is not returned as a possible anagram. I believe this is because of line 28-31 (below):
python for letter in word: if word_letter_map[letter] <= name_letter_map[letter]: test += letter if Counter(test) == word_letter_map: anagrams.append(word)
The test string only receives a singular letter, rather than the value attached to the letter, making it impossible to have an anagram where a single letter appears more than once in the word. Once we get to the second if statement, we only have a string (test) comprised of letters that all have a frequency of 1, limiting the words we can utilize.
Was this an intentional decision or am I possibly misinterpreting something?
I updated the function with the following logic (seems to work, but curious if you see an error):
python for letter in word: if word_letter_map[letter] <= name_letter_map[letter]: test += letterword_letter_map[letter] if Counter(test) == word_letter_map: anagrams.append(word) Now it will add a letter to the string 'test', but will add that letter the number of times it exists in the word of interest. On some basic testing it seems to work, but I am curious if I am overlooking something here.
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dwanneruchi
@rlvaugh this book is great! Really enjoying it!
It looks like find_anagrams() is only able to produce anagrams composed of unique letters (i.e. n letters with n distinct letters).
For example, my name is david. When I run this function the word "dad" is not returned as a possible anagram. I believe this is because of line 28-31 (below):
The test string only receives a singular letter, rather than the value attached to the letter, making it impossible to have an anagram where a single letter appears more than once in the word. Once we get to the second if statement, we only have a string (test) comprised of letters that all have a frequency of 1, limiting the words we can utilize.
Was this an intentional decision or am I possibly misinterpreting something?
I updated the function with the following logic (seems to work, but curious if you see an error):
Now it will add a letter to the string 'test', but will add that letter * the number of times it exists in the word of interest. On some basic testing it seems to work, but I am curious if I am overlooking something here. Do you think this is a good modification?