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@cibofdevs
Last active December 9, 2023 18:08
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# 1. The dictionary Junior shows a schedule for a junior year semester.
# The key is the course name and the value is the number of credits.
# Find the total number of credits taken this semester and assign it to the variable credits.
# Do not hardcode this – use dictionary accumulation!
Junior = {'SI 206':4, 'SI 310':4, 'BL 300':3, 'TO 313':3, 'BCOM 350':1, 'MO 300':3}
credits = 0
for v in Junior.values():
credits += v
print(credits)
# 2. Create a dictionary, freq, that displays each character in string str1 as the key and its frequency as the value.
from collections import Counter
str1 = "peter piper picked a peck of pickled peppers"
freq = Counter(str1)
for i in str1:
print(i, freq[i])
# 3. Provided is a string saved to the variable name s1.
# Create a dictionary named counts that contains each letter in s1 and the number of times it occurs.
s1 = "hello"
def char_frequency(s1):
dict = {}
for n in s1:
keys = dict.keys()
if n in keys:
dict[n] += 1
else:
dict[n] = 1
return dict
counts = char_frequency(s1)
print(counts)
# 4. Create a dictionary, freq_words, that displays each word in string str1 as the key and its frequency as the value.
str1 = "I wish I wish with all my heart to fly with dragons in a land apart"
def word_count(str):
counts = dict()
words = str.split()
for word in words:
if word in counts:
counts[word] += 1
else:
counts[word] = 1
return counts
freq_words = word_count(str1)
print(freq_words)
# 5. Create a dictionary called wrd_d from the string sent,
# so that the key is a word and the value is how many times you have seen that word.
sent = "Singing in the rain and playing in the rain are two entirely different situations but both can be good"
def word_count(str):
counts = dict()
words = str.split()
for word in words:
if word in counts:
counts[word] += 1
else:
counts[word] = 1
return counts
wrd_d = word_count(sent)
print(wrd_d)
# 6. Create the dictionary characters that shows each character from the string sally and its frequency.
# Then, find the most frequent letter based on the dictionary. Assign this letter to the variable best_char.
sally = "sally sells sea shells by the sea shore"
characters = {}
for i in sally:
characters[i]=characters.get(i,0)+1
sorted(characters.items(), key=lambda x: x[1])
best_char = sorted(characters.items(), key=lambda x: x[1])[-1][0]
# 7. Do the same as above but now find the least frequent letter.
# Create the dictionary characters that shows each character from string sally and its frequency.
# Then, find the least frequent letter in the string and assign the letter to the variable worst_char.
sally = "sally sells sea shells by the sea shore and by the road"
characters = {}
for i in sally:
characters[i]=characters.get(i,0)+1
sorted(characters.items(), key=lambda x: x[1])
worst_char = sorted(characters.items(), key=lambda x: x[1])[-13][0]
# 9. Create a dictionary called low_d that keeps track of all the characters in the string p
# and notes how many times each character was seen. Make sure that there are no repeats of characters as keys,
# such that “T” and “t” are both seen as a “t” for example.
import collections
p = p = "Summer is a great time to go outside. You have to be careful of the sun though because of the heat."
low_d = collections.defaultdict(int)
for c in p:
low_d[c] += 1
for c in sorted(low_d, key=low_d.get, reverse=True):
if low_d[c] > 1:
print('%s %d' % (c, low_d[c]))
@xojamurodtuit
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help #8

@xojamurodtuit
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help #9

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