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Kevin Selagea
2,161 PointsI have no idea what I'm doing
Here's my code. I have no idea what exactly the challenge is asking me to do, but I have a decent idea. Can any of you guys help me out? Thanks in advance.
# E.g. word_count("I do not like it Sam I Am") gets back a dictionary like:
# {'i': 2, 'do': 1, 'it': 1, 'sam': 1, 'like': 1, 'not': 1, 'am': 1}
# Lowercase the string to make it easier.
string_dict = "{i} {am} {that} {i} {am}"
split_string = string_dict.split()
def word_count(string):
string = string.lower()
new_string = string.split(' ')
final = {}
for key in new_string:
if key in final:
final[key] += 1
else:
final[key] = 1
return final
1 Answer
Stuart McIntosh
Python Web Development Techdegree Graduate 22,874 PointsHi there, I am sure there are better ways to do this - however here was my answer
def word_count(my_str):
t_list = my_str.lower().split() # create a list to iterate through
n_dict = {} # this will be the returned dictionary
t_dict = {} # temporay dict to hold the key and value
for x in t_list: # iterate through the list
x_ctr = 0 # set a counter to 0 this will hold the value
for y in t_list:
if x == y: # if we find a similar value add 1 to the count
x_ctr += 1
t_dict = {x : x_ctr} # create a temp dict with the key and value
n_dict.update(t_dict) # append the temp dict to the final dict
return n_dict
if __name__ == "__main__":
print(word_count('I do not like it Sam I am'))
Hopefully this is easily readable, if not please ask me a question