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Algorithm to get the excel-like column name of a number
...etc)...
function getNameFromNumber($num) {
$numeric = $num % 26;
$letter = chr(65 + $numeric);
$num2 = intval($num / 26);
if ($num2 > 0) {
return getNameFromNumber($num2 - 1) . $letter;
} else {
return $letter;
}
}
And if you want it one indexed (1 == A,...
How to generate keyboard events in Python?
...ist-of-mac-virtual-key-codes
def toKeyCode(c):
shiftKey = False
# Letter
if c.isalpha():
if not c.islower():
shiftKey = True
c = c.lower()
if c in shiftChars:
shiftKey = True
c = shiftChars[c]
if c in keyCodeMap:
keyCode =...
Efficiently replace all accented characters in a string?
...iacritics (str) {
var defaultDiacriticsRemovalMap = [
{'base':'A', 'letters':/[\u0041\u24B6\uFF21\u00C0\u00C1\u00C2\u1EA6\u1EA4\u1EAA\u1EA8\u00C3\u0100\u0102\u1EB0\u1EAE\u1EB4\u1EB2\u0226\u01E0\u00C4\u01DE\u1EA2\u00C5\u01FA\u01CD\u0200\u0202\u1EA0\u1EAC\u1EB6\u1E00\u0104\u023A\u2C6F]/g},
...
Reorder levels of a factor without changing order of values
...
Use the levels argument of factor:
df <- data.frame(f = 1:4, g = letters[1:4])
df
# f g
# 1 1 a
# 2 2 b
# 3 3 c
# 4 4 d
levels(df$g)
# [1] "a" "b" "c" "d"
df$g <- factor(df$g, levels = letters[4:1])
# levels(df$g)
# [1] "d" "c" "b" "a"
df
# f g
# 1 1 a
# 2 2 b
# 3 3 c
# 4 4 d
...
What's the difference between `raw_input()` and `input()` in Python 3?
What is the difference between raw_input() and input() in Python 3?
6 Answers
6
...
Tying in to Django Admin's Model History
...ib.admin.models import LogEntry, ADDITION
LogEntry.objects.log_action(
user_id = request.user.pk,
content_type_id = ContentType.objects.get_for_model(object).pk,
object_id = object.pk,
object_repr = force_unicode(object),
action_flag = ADDITION
)
where o...
Generate a random letter in Python
Is there a way to generate random letters in Python (like random.randint but for letters)? The range functionality of random.randint would be nice but having a generator that just outputs a random letter would be better than nothing.
...
Algorithm for classifying words for hangman difficulty levels as “Easy”,“Medium”, or “Hard”
...rategy for the solver would be to base their decisions on the frequency of letters in English, or on the frequency of words in some corpus. This is a seductive idea, but it's not quite right. The solver does best if it accurately models the distribution of words chosen by the setter, and a human set...
How to create an object for a Django model with a many to many field?
...()
sample_id = sample_object.id
sample_object.users.through.objects.create(user_id=1, sample_id=sample_id)
sample_object.users.through.objects.create(user_id=2, sample_id=sample_id)
This will work because we already know that the 'users' list is empty, so we can create mindlessly.
...
How to check if any flags of a flag combination are set?
...
If you want to know if letter has any of the letters in AB you must use the AND & operator. Something like:
if ((letter & Letters.AB) != 0)
{
// Some flag (A,B or both) is enabled
}
else
{
// None of them are enabled
}
...