I want to create ID for each element inserted in an empty dictionary then write it in a file as in the picture below. But it doesn't work. Any help to fix it?
dict ={}
ids = 0
line_count = 0
fhand = input('Enter the file name:')
fname = open(fhand,'a+')
for line in fname:
if line.split() == []:
ids = 1
else:
line_count +=1
ids = line_count +1
n = int(input('How many colors do you want to add?'))
for i in range (0,n):
dict['ID:'] = ids + 1
dict['Color:'] = input('Enter the color:')
for key,value in dict.items():
s = str(key)+' '+str(value)+'\n'
fname.write(s)
fname.close()
print('Done!') ```
Output should be:
ID : 1
Color: red
ID : 2
Color : rose
ID : 3
Color : blue
Not sure if I got what you meant but...
A dictionary is made of <key, value> pairs.
Let`s suppose you have a dictionary:
thisdict = {
"brand": "Ford",
"model": "Mustang",
"year": 1964
}
If you want to create a ID for each key (for a specific reason) you could use a for loop like:
for key in thisdict.keys():
createIdFunction(key)
And have a createIdFunction which is going to assign a ID based on whatever you want.
Suggestion: Dictionaries can only hold unique keys, so maybe you could use their own keys as ID.
However if your dictionary is empty, there would be no reason to have a ID for that key, right?
You mean your id is not increased ? I think you did not reassign variable "ids" in loop, you may modify code as below:
dict ={}
ids = 0
line_count = 0
fhand = input('Enter the file name:')
fname = open(fhand,'a+')
for line in fname:
if line.split() == []:
ids = 1
else:
line_count +=1
ids = line_count +1
n = int(input('How many colors do you want to add?'))
for i in range (0,n):
ids += 1 # modified
dict['ID:'] = ids # modified
dict['Color:'] = input('Enter the color:')
for key,value in dict.items():
s = str(key)+' '+str(value)+'\n'
fname.write(s)
fname.close()
Related
This is how my code looks in python. in the line where df = get_data_df(id,start_at) , instead of defining id one by one, I would like for my program to iterate over id and use it in the program below. please help me with how to iterate over the dictionary (id) and use it in the while loop.
id= {'O': 6232,
'S': 5819,
'S': 5759,
'R': 6056,
'M': 6145,}
whole_df = pd.DataFrame()
start_at = int(datetime(2020,8,1,6,0,0,0, pytz.UTC).timestamp() * 1e6)
while True:
df = get_data_df(id,start_at)
if df.shape[0] <= 1:
break
else:
whole_df = whole_df.append(df)
last_timestamp = whole_df.last_valid_index().timestamp()
start_at = int(last_timestamp * 1e6)
#print(whole_df)
for key in id:
print(key)
Is one way you could do it. Or you Could do it this way
i = 0
while True:
list(id)[i]
i += 1
By Just itterating a index and grabbing a value at each point from that Index
There a multiple ways of iterating over a python dictionary using a for loop:
for key in your_dict:
value = your_dict[key]
print(value)
for value in your_dict.values():
print(value)
for key, value in your_dict.items():
print(key, '=', value)
If you realy want a while loop:
keys = your_dict.keys()
i = 0
while i < len(keys):
value = your_dict[keys[i]]
print(key, '=', value)
Given the names and grades for each student in a Physics class of students, store them in a nested list and print the name(s) of any student(s) having the second lowest grade.
Note: If there are multiple students with the same grade, order their names alphabetically and print each name on a new line.
Input Format
The first line contains an integer, , the number of students.
The subsequent lines describe each student over lines; the first line contains a student's name, and the second line contains their grade.
Constraints
There will always be one or more students having the second lowest grade.
Output Format
Print the name(s) of any student(s) having the second lowest grade in Physics; if there are multiple students, order their names alphabetically and print each one on a new line.
This is my code:
list = []
for _ in range(int(input())):
name = input()
score = float(input())
new = [name, score]
list.append(new)
def snd_highest(val):
return val[1]
list.sort(key = snd_highest)
list.sort()
value = list[1]
grade = value[1]
for a,b in list:
if b == grade:
print (a)
This is the test case:
4
Rachel
-50
Mawer
-50
Sheen
-50
Shaheen
51
And the expected output is Shaheen but i got the other 3.
Please explain.
To find the second lowest value, you have actually just sorted your list in ascending order and just taken the second value in the list by using the below code
value = list[1]
grade = value[1]
Imagine this is your list after sorting:
[['Sheen', 50.0], ['mawer', 50.0], ['rachel', 50.0], ['shaheen', 51.0]]
According to value = list[1], the program chooses "value = ['mawer', 50.0]".
Then the rest of your program takes the grade from this value and outputs the corresponding name, that's why this isn't working as per what you need, you need to write logic to find the lowest value and then find the second lowest, this current program just assumes the lowest value is in the second position in the list.
Try doing this
if __name__ == '__main__':
students = []
for _ in range(int(input())):
name = input()
score = float(input())
new = [name, score]
students.append(new)
def removeMinimum(oldlist):
oldlist = sorted(oldlist, key=lambda x: x[1])
min_ = min(students, key=lambda x: x[1])
newlist = []
for a in range(0, len(oldlist)):
if min_[1] != oldlist[a][1]:
newlist.append(oldlist[a])
return newlist
students = removeMinimum(students);
# find the second minimum value
min_ = min(students, key=lambda x: x[1])
# sort alphabetic order
students = sorted(students, key=lambda x: x[0])
for a in range(0, len(students)):
if min_[1] == students[a][1]:
print(students[a][0])
I hope this may help you to pass all your test cases. Thank you.
# These functions will be used for sorting
def getSecond(ele):
return ele[1]
def getFirst(ele):
return ele[0]
studendList = []
sortedList = []
secondLowestStudents = []
# Reading input from STDIN and saving in nested list [["stud1": <score>], ["stud2", <score>]]
for _ in range(int(input())):
name = input()
score = float(input())
studendList.append([name, score])
# sort the list by score and save it in a new list studendList (remove the duplicate score as well - see, if x[1] not in sortedList)
studendList.sort(key=getSecond)
[sortedList.append(x[1]) for x in studendList if x[1] not in sortedList]
# Get the second lowest grade
secondLowest = sortedList[1]
# Now sort the origin list by the name fetch the student list having the secondLowest grade
studendList.sort(key=getFirst)
[secondLowestStudents.append(x[0]) for x in studendList if x[1] == secondLowest]
# Print the student's name having second-lowest grade
for st in secondLowestStudents:
print(st)
I'm fairly new to Python but I haven't found the answer to this particular problem.
I am writing a simple recommendation program and I need to have a dictionary where cuisine is a key and name of a restaurant is a value. There are a few instances where I have to split a string of a few cuisine names and make sure all other restaurants (values) which have the same cuisine get assigned to the same cuisine (key). Here's a part of a file:
Georgie Porgie
87%
$$$
Canadian, Pub Food
Queen St. Cafe
82%
$
Malaysian, Thai
Mexican Grill
85%
$$
Mexican
Deep Fried Everything
52%
$
Pub Food
so it's just the first and the last one with the same cuisine but there are more later in the file.
And here is my code:
def new(file):
file = "/.../Restaurants.txt"
d = {}
key = []
with open(file) as file:
lines = file.readlines()
for i in range(len(lines)):
if i % 5 == 0:
if "," not in lines[i + 3]:
d[lines[i + 3].strip()] = [lines[i].strip()]
else:
key += (lines[i + 3].strip().split(', '))
for j in key:
if j not in d:
d[j] = [lines[i].strip()]
else:
d[j].append(lines[i].strip())
return d
It gets all the keys and values printed but it doesn't assign two values to the same key where it should. Also, with this last 'else' statement, the second restaurant is assigned to the wrong key as a second value. This should not happen. I would appreciate any comments or help.
In the case when there is only one category you don't check if the key is in the dictionary. You should do this analogously as in the case of multiple categories and then it works fine.
I don't know why you have file as an argument when you have a file then overwritten.
Additionally you should make 'key' for each result, and not += (adding it to the existing 'key'
when you check if j is in dictionary, clean way is to check if j is in the keys (d.keys())
def new(file):
file = "/.../Restaurants.txt"
d = {}
key = []
with open(file) as file:
lines = file.readlines()
for i in range(len(lines)):
if i % 5 == 0:
if "," not in lines[i + 3]:
if lines[i + 3] not in d.keys():
d[lines[i + 3].strip()] = [lines[i].strip()]
else:
d[lines[i + 3]].append(lines[i].strip())
else:
key = (lines[i + 3].strip().split(', '))
for j in key:
if j not in d.keys():
d[j] = [lines[i].strip()]
else:
d[j].append(lines[i].strip())
return d
Normally, I find that if you use names for the dictionary keys, you may have an easier time handling them later.
In the example below, I return a series of dictionaries, one for each restaurant. I also wrap the functionality of processing the values in a method called add_value(), to keep the code more readable.
In my example, I'm using codecs to decode the value. Although not necessary, depending on the characters you are dealing with it may be useful. I'm also using itertools to read the file lines with an iterator. Again, not necessary depending on the case, but might be useful if you are dealing with really big files.
import copy, itertools, codecs
class RestaurantListParser(object):
file_name = "restaurants.txt"
base_item = {
"_type": "undefined",
"_fields": {
"name": "undefined",
"nationality": "undefined",
"rating": "undefined",
"pricing": "undefined",
}
}
def add_value(self, formatted_item, field_name, field_value):
if isinstance(field_value, basestring):
# handle encoding, strip, process the values as you need.
field_value = codecs.encode(field_value, 'utf-8').strip()
formatted_item["_fields"][field_name] = field_value
else:
print 'Error parsing field "%s", with value: %s' % (field_name, field_value)
def generator(self, file_name):
with open(file_name) as file:
while True:
lines = tuple(itertools.islice(file, 5))
if not lines: break
# Initialize our dictionary for this item
formatted_item = copy.deepcopy(self.base_item)
if "," not in lines[3]:
formatted_item['_type'] = lines[3].strip()
else:
formatted_item['_type'] = lines[3].split(',')[1].strip()
self.add_value(formatted_item, 'nationality', lines[3].split(',')[0])
self.add_value(formatted_item, 'name', lines[0])
self.add_value(formatted_item, 'rating', lines[1])
self.add_value(formatted_item, 'pricing', lines[2])
yield formatted_item
def split_by_type(self):
d = {}
for restaurant in self.generator(self.file_name):
if restaurant['_type'] not in d:
d[restaurant['_type']] = [restaurant['_fields']]
else:
d[restaurant['_type']] += [restaurant['_fields']]
return d
Then, if you run:
p = RestaurantListParser()
print p.split_by_type()
You should get:
{
'Mexican': [{
'name': 'Mexican Grill',
'nationality': 'undefined',
'pricing': '$$',
'rating': '85%'
}],
'Pub Food': [{
'name': 'Georgie Porgie',
'nationality': 'Canadian',
'pricing': '$$$',
'rating': '87%'
}, {
'name': 'Deep Fried Everything',
'nationality': 'undefined',
'pricing': '$',
'rating': '52%'
}],
'Thai': [{
'name': 'Queen St. Cafe',
'nationality': 'Malaysian',
'pricing': '$',
'rating': '82%'
}]
}
Your solution is simple, so it's ok. I'd just like to mention a couple of ideas that come to mind when I think about this kind of problem.
Here's another take, using defaultdict and split to simplify things.
from collections import defaultdict
record_keys = ['name', 'rating', 'price', 'cuisine']
def load(file):
with open(file) as file:
data = file.read()
restaurants = []
# chop up input on each blank line (2 newlines in a row)
for record in data.split("\n\n"):
fields = record.split("\n")
# build a dictionary by zipping together the fixed set
# of field names and the values from this particular record
restaurant = dict(zip(record_keys, fields))
# split chops apart the type cuisine on comma, then _.strip()
# removes any leading/trailing whitespace on each type of cuisine
restaurant['cuisine'] = [_.strip() for _ in restaurant['cuisine'].split(",")]
restaurants.append(restaurant)
return restaurants
def build_index(database, key, value):
index = defaultdict(set)
for record in database:
for v in record.get(key, []):
# defaultdict will create a set if one is not present or add to it if one does
index[v].add(record[value])
return index
restaurant_db = load('/var/tmp/r')
print(restaurant_db)
by_type = build_index(restaurant_db, 'cuisine', 'name')
print(by_type)
Python 3
Hello guys I'm a python beginner studying dictionary now
below is what I have learned so far how to save list into a file
and count items in list like below.
class item:
name = None
score = None
def save_list_in_file(file_name:str, L:[item]):
f = open(file_name,'w')
for it in L:
f.write(it.name + "" + str(it.score))
f.close()
def count_item_in_list(L:[item])->int:
n = 0
for it in L:
if it.score >= 72:
n += 1
return n
and I'm not sure if using dictionary is same way as I use in list
for example:
def save_dict_to_file(file_name:str, D:{item}):
f = open(file_name,'w')
for it in D:
f.write(it.name + "" + str(it.score))
f.close()
def count_item_in_dict(D:{item})->int:
n = 0
for it in D:
if it.score <= 72:
n += 1
return n
will be correct? i thought dict would be different than using a list.
Thanks for any comment!
You can't use a dictionary the same way as using a list.
A list is defined as a sequence of elements. So when you have the list:
L=['D','a','v','i','d']
You can loop it like this:
for it in L:
print(it)
And it will print:
D
a
v
i
d
Instead, a dictionary is a group of tuples of two elements where one is the key and the second is the value. So for example you have a Dictonary like this:
D = {'firstletter' : 'D', 'secondletter': 'a', 'thirdletter' : 'v' }
And when you loop it like a list:
for it in L:
print(it)
it will print only the keys:
firstletter
secondletter
thirdletter
so in order to obtain the values you have to print it like this:
for it in D:
print(D[it])
that will display this result:
D
a
v
If you need more information you can check de documentation for dictionary :)
Python 3 documentation of Data Structures
I'm new to Python and programming in general and need a little help with this (partially finished) function. It's calling a text file with a bunch of rows of comma delimited data (age, salary, education and so on). However, I've run into a problem from the outset. I don't know how to return the results.
My aim is to create dictionaries for each category and for each row to be sorted and tallied.
e.g. 100 people over 50, 200 people under 50 and so on.
Am I in the correct ball park?
file = "adultdata.txt"
def make_data(file):
try:
f = open(file, "r")
except IOError as e:
print(e)
return none
large_list = []
avg_age = 0
row_count_under50 = 0
row_count_over50 = 0
#create 2 dictionaries per category
employ_dict_under50 = {}
employ_dict_over50 = {}
for row in f:
edited_row = row.strip()
my_list = edited_row.split(",")
try:
#Age Category
my_list[0] = int(my_list[0])
#Work Category
if my_list[-1] == " <=50K":
if my_list[1] in employ_dict_under50:
employ_dict_under50[my_list[1]] += 1
else:
employ_dict_under50[my_list[1]] = 1
row_count_u50 += 1
else:
if my_list[1] in emp_dict_o50:
employ_dict_over50[my_list[1]] += 1
else:
employ_dict_over50[my_list[1]] = 1
row_count_o50 += 1
# Other categories here
print(my_list)
#print(large_list)
#return
# Ignored categories here - e.g. my_list[insert my list numbers here] = None
I do not have access to your file but I had a go at correcting most of the errors you had in your code.
These are a list of the mistakes I found in your code:
your function make_data is essentially useless and is out of scope. You need to remove it entirely
When using a file object f, you need to use readline to extract data from the file.
It is also best to use a with statement when using IO resources like files
You had numerous variables which were badly named in the inner loop and did not exist
You declared a try in the inner loop without a catch. You can remove the try because you are not trying to catch any Error
You have some very basic errors which are related to general programming, can I assume your new to this? If thats the case then you should probably follow some more beginner tutorials online until you get a grasp of what commands you need to use to perform basic tasks.
Try compare your code to this and see if you can understand what i'm trying to say:
file = "adultdata.txt"
large_list = []
avg_age = 0
row_count_under50 = 0
row_count_over50 = 0
#create 2 dictionaries per category
employ_dict_under50 = {}
employ_dict_over50 = {}
with open(file, "r") as f:
row = f.readline()
edited_row = row.strip()
my_list = edited_row.split(",")
#Age Category
my_list[0] = int(my_list[0])
#Work Category
if my_list[-1] == " <=50K":
if my_list[1] in employ_dict_under50:
employ_dict_under50[my_list[1]] += 1
else:
employ_dict_under50[my_list[1]] = 1
row_count_under50 += 1
else:
if my_list[1] in employ_dict_over50:
employ_dict_over50[my_list[1]] += 1
else:
employ_dict_over50[my_list[1]] = 1
row_count_over50 += 1
# Other categories here
print(my_list)
#print(large_list)
#return
I cannot say for certain if this code will work or not without your file but it should give you a head start.