I want to add numbers to list of strings so I can keep count of number of items.
Assume I have a list like this:
list = ['a', 'b', 'A', 'b', 'a', 'C', 'D', 'd']
I want to assign a number to each string irrespective of capital or small letters, Here's the output I'm looking for,
list = ['a_1', 'b_1', 'A_2', 'b_2', 'a_3', 'C_1', 'D_1', 'd_2']
This is what I've tried but I'm not getting the correct output
list = [j+f'_{i}' for i, j in enumerate(lst, 1)]
You can keep track of how many times you have seen a number in a dictionary, and update the count whenever you see the letter, and use the last count to append to the character.
from collections import defaultdict
def label(lst):
dct = defaultdict(int)
output = []
#Iterate through the list
for item in lst:
char = item.lower()
#Update dictionary
dct[char] += 1
#Create the list of characters with count appended
output.append(f'{item}_{dct[char]}')
return output
print(label(['a', 'b', 'A', 'b', 'a', 'C', 'D', 'd']))
The output will be
['a_1', 'b_1', 'A_2', 'b_2', 'a_3', 'C_1', 'D_1', 'd_2']
Also don't use list as a variable name, since it's a reserved python builtin name.
You can keep track of the counts using a dictionary:
newlist = []
counts = {}
for element in oldlist:
if element.casefold() in counts:
counts[element.casefold()] += 1
else:
counts[element.casefold()] = 1
newlist.append(f'{element}_{counts[element.casefold()]}')
counts = defaultdict(int), you could just increment the counter and don't need to write the if-else-statement (see answer of Devesh for example)You can use a Counter for this to keep track of the number of times the item has been added, like:
from collections import Counter
def labelize(iterable):
c = Counter()
for item in iterable:
sc = item.casefold()
c[sc] += 1
yield '{}_{}'.format(item, c[sc])We can then label with:
>>> list(labelize(['a', 'b', 'A', 'b', 'a', 'C', 'D', 'd']))
['a_1', 'b_1', 'A_2', 'b_2', 'a_3', 'C_1', 'D_1', 'd_2']
We can generalize the above by using an arbitrary mapping function, like:
def labelize(iterable, key=lambda x: x):
c = Counter()
for item in iterable:
ci = key(item)
c[ci] += 1
yield '{}_{}'.format(item, c[ci])The equivalent here would be:
>>> list(labelize(['a', 'b', 'A', 'b', 'a', 'C', 'D', 'd'], str.casefold))
['a_1', 'b_1', 'A_2', 'b_2', 'a_3', 'C_1', 'D_1', 'd_2']
Note: please do not use
str.lowerorstr.upperfor case matching. Certain cultures have specific rules for case matching. For example in German, the eszett ß [wiki] has as uppercase'SS', by using.tolower(), the two would be different). Thestr.casefold[Python-doc] is designed to map strings to a value that can be used for case-insensitive matching. See for more information 3.13 Default Case Algorithms of the Unicode 9 standard.
Here is another way:
original = ['a', 'b', 'A', 'b', 'a', 'C', 'D', 'd']
lowered_reversed = [i.lower() for i in original][::-1]
for i in range(len(lowered_reversed)):
count = lowered_reversed.count(lowered_reversed[i])
lowered_reversed[i] = lowered_reversed[i] + "_" + str(count)
new_list = lowered_reversed[::-1]
for i in range(len(original)):
if original[i].isupper():
new_list[i] = new_list[i].upper()
print(new_list)
Output:
['a_1', 'b_1', 'A_2', 'b_2', 'a_3', 'C_1', 'D_1', 'd_2']
range. you even called the variable i which is usually for indices
list1? should be original
listsince it's a python reserved name[a_1, b_2, a_3, b_4, a_5, c_6, D_7, d_8]a = A?