Redundant copies of a dictionary as Python list - python-3.x

I am trying to make redundant copies of a dictionary as a list, but update one of the keys. Initially the list has just one item, say, like this:
l1 = [{'case':1}]
I ran the below loop:
for i in range(1,4):
l1.append(l1[i-1])
l1[i]['case'] += 1
I expected the output to be: [{'case':1},{'case':2},{'case':3},{'case':4}]
The actual output is like this: [{'case':4},{'case':4},{'case':4},{'case':4}]
How can I get the output as expected?

Python keeps a reference to the original dict, so if you alter the dict with l1[i]['case'] += 1 it alters every dict in your list since they are references to the same object (in other words they are the same dict).
In order to make them independent you have to call dict.copy(), so if you change your code to the following it will work:
l1 = [{'case':1}]
for i in range(1,4):
l1.append(l1[i-1].copy())
l1[i]["case"] += 1
print(l1)

Related

Iterating thru a not so ordinary Dictionary in python 3.x

Maybe it is ordinary issue regarding iterating thru a dict. Please find below imovel.txt file, whose content is as follows:
{'Andar': ['primeiro', 'segundo', 'terceiro'], 'Apto': ['101','201','301']}
As you can see this is not a ordinary dictionary, with a key value pair; but a key with a list as key and another list as value
My code is:
#/usr/bin/python
def load_dict_from_file():
f = open('../txt/imovel.txt','r')
data=f.read()
f.close()
return eval(data)
thisdict = load_dict_from_file()
for key,value in thisdict.items():
print(value)
and yields :
['primeiro', 'segundo', 'terceiro'] ['101', '201', '301']
I would like to print a key,value pair like
{'primeiro':'101, 'segundo':'201', 'terceiro':'301'}
Given such txt file above, is it possible?
You should use the builtin json module to parse but either way, you'll still have the same structure.
There are a few things you can do.
If you know both of the base key names('Andar' and 'Apto') you can do it as a one line dict comprehension by zipping the values together.
# what you'll get from the file
thisdict = {'Andar': ['primeiro', 'segundo', 'terceiro'], 'Apto': ['101','201','301']}
# One line dict comprehension
newdict = {key: value for key, value in zip(thisdict['Andar'], thisdict['Apto'])}
print(newdict)
If you don't know the names of the keys, you could call next on an iterator assuming they're the first 2 lists in your structure.
# what you'll get from the file
thisdict = {'Andar': ['primeiro', 'segundo', 'terceiro'], 'Apto': ['101','201','301']}
# create an iterator of the values since the keys are meaningless here
iterator = iter(thisdict.values())
# the first group of values are the keys
keys = next(iterator, None)
# and the second are the values
values = next(iterator, None)
# zip them together and have dict do the work for you
newdict = dict(zip(keys, values))
print(newdict)
As other folks have noted, that looks like JSON, and it'd probably be easier to parse it read through it as such. But if that's not an option for some reason, you can look through your dictionary this way if all of your lists at each key are the same length:
for i, res in enumerate(dict[list(dict)[0]]):
ith_values = [elem[i] for elem in dict.values()]
print(ith_values)
If they're all different lengths, then you'll need to put some logic to check for that and print a blank or do some error handling for looking past the end of the list.

Python equivalent of array of structs from MATLAB

I am familiar with the struct construct from MATLAB, specifically array of structs. I am trying to do that with dictionary in Python. Say I have a initialized a dictionary:
samples = {"Name":"", "Group":"", "Timeseries":[],"GeneratedFeature":[]}
and I am provided with another dictionary called fileList whose keys are group names and each value is a tuples of file-paths. Each file path will generate one sample in samples by populating the Timeseries item. Further some processing will make GeneratedFeature. The name part will be determined by the filepath.
Since I don't know the contents of fileList a priori, in MATLAB if samples were a struct and fileList just a cell array:
fileList={{'Group A',{'filepath1','filepath2'}};{'Group B',{'filepath1', 'filepath2'}}}
I would just set a counter k=1 and run a for loop (with a different index) and do something like:
k=1;
for i=1:numel(fileList)
samples(k).Group=fileList{i}{1};
for j=1:numel(fileList{i}{2})
samples(k).Name=makeNameFrom(fileList{1}{2}{j})
.
.
end
k=k+1
end
But I don't know how to do this in python. I know I can keep the two for loop approach with
for (group, samples) in fileList:
for sample in samples:
But how to tell python that samples is allowed to be an array/list? Is there a more pythonic approach than doing for loop?
You could store your dictionary itself in a list and simply append new dictionaries in every iteration of the loop:
samplelist = []
samplelist.append(samples.copy()) % dictionary copy needed when duplicating
Accessing the elements in the list would then work as follows (For example the 'Name' field of the i-th sample):
samples_i_name = samplelist[i]["Name"]
A list of all names would be accessible by a simple list comprehension:
namelist = [samplelist[i]["Name"] for i in range(len(samplelist))]

List, tuples or dictionary, differences and usage, How can I store info in python

I'm very new in python (I usually write in php). I want to understand how to store information in an associative array, and if you can explain me whats the difference of "tuples", "arrays", "dictionary" and "list" will be wonderful (I tried to read different source but I still not caching it).
So This is my code:
#!/usr/bin/python3.4
import csv
import string
nidless_keys = dict()
nidless_keys = ['test_string1','test_string2'] #this contain the string to
# be searched in linesreader
data = {'type':[],'id':[]} #here I want to store my information
with open('path/to/csv/file.csv',newline="") as csvfile:
linesreader = csv.reader(csvfile,delimiter=',',quotechar="|")
for row in linesreader: #every line in this csv have a url like
#www.test.com/?test_string1&id=123456
current_row_string = str(row)
for needle in nidless_keys:
current_needle = str(needle)
if current_needle in current_row_string:
data[current_needle[current_row_string[-8:]]) += 1 # also I
#need to count per every id how much rows there are.
In conclusion:
my_data_stored = [current_needle][current_row_string[-8]]
current_row_string[-8] is a url which the last 8 digit of the url is an ID.
So the array should looks like this at the end of the script:
test_string1 = 123456 = 20
= 256468 = 15
test_string2 = 123155 = 10
Edit 1:
Which type I need here to store the information?
Can you tell me how to resolve this script?
It seems you want to count how many times an ID in combination with a test string occurs.
There can be multiple ID/count combinations associated with every test string.
This suggests that you should use a dictionary indexed by the test strings to store the results. In that dictionary I would suggest to store collections.Counter objects.
This way, you would have to add a special case when a key in the results dictionary isn't found to add an empty Counter. This is a common problem, so there is a specialized form of dictionary in the collections module called defaultdict.
import collections
import csv
# Using a tuple for the keys so it cannot be accidentally modified
keys = ('test_string1', 'test_string2')
result = collections.defaultdict(collections.Counter)
with open('path/to/csv/file.csv',newline="") as csvfile:
linesreader = csv.reader(csvfile,delimiter=',',quotechar="|")
for row in linesreader:
for key in keys:
if key in row:
id = row[-6:] # ID's are six digits in your example.
# The first index is into the dict, the second into the Counter.
result[key][id] += 1
There is an even easier way, by using regular expressions.
Since you seem to treat every row in a CSV file as a string, there is little need to use the CSV reader, so I'll just read the whole file as text.
import re
with open('path/to/csv/file.csv') as datafile:
text = datafile.read()
pattern = r'\?(.*)&id=(\d+)'
The pattern is a regular expression. This is a large topic in and of itself, so I'll only cover briefly what it does. (You might also want to check out the relevant HOWTO) At first glance it looks like complete gibberish, but it is actually a complete language.
In looks for two things in a line. Anything between ? and &id=, and a sequence of digits after &id=.
I'll be using IPython to give an example.
(If you don't know it, check out IPython. It is great for trying things and see if they work.)
In [1]: import re
In [2]: pattern = r'\?(.*)&id=(\d+)'
In [3]: text = """www.test.com/?test_string1&id=123456
....: www.test.com/?test_string1&id=123456
....: www.test.com/?test_string1&id=234567
....: www.test.com/?foo&id=234567
....: www.test.com/?foo&id=123456
....: www.test.com/?foo&id=1234
....: www.test.com/?foo&id=1234
....: www.test.com/?foo&id=1234"""
The text variable points to the string which is a mock-up for the contents of your CSV file.
I am assuming that:
every URL is on its own line
ID's are a sequence of digits.
If these assumptions are wrong, this won't work.
Using findall to extract every match of the pattern from the text.
In [4]: re.findall(pattern, test)
Out[4]:
[('test_string1', '123456'),
('test_string1', '123456'),
('test_string1', '234567'),
('foo', '234567'),
('foo', '123456'),
('foo', '1234'),
('foo', '1234'),
('foo', '1234')]
The findall function returns a list of 2-tuples (that is key, ID pairs). Now we just need to count those.
In [5]: import collections
In [6]: result = collections.defaultdict(collections.Counter)
In [7]: intermediate = re.findall(pattern, test)
Now we fill the result dict from the list of matches that is the intermediate result.
In [8]: for key, id in intermediate:
....: result[key][id] += 1
....:
In [9]: print(result)
defaultdict(<class 'collections.Counter'>, {'foo': Counter({'1234': 3, '123456': 1, '234567': 1}), 'test_string1': Counter({'123456': 2, '234567': 1})})
So the complete code would be:
import collections
import re
with open('path/to/csv/file.csv') as datafile:
text = datafile.read()
result = collections.defaultdict(collections.Counter)
pattern = r'\?(.*)&id=(\d+)'
intermediate = re.findall(pattern, test)
for key, id in intermediate:
result[key][id] += 1
This approach has two advantages.
You don't have to know the keys in advance.
ID's are not limited to six digits.
A brief summary of the python data types you mentioned:
A dictionary is an associative array, aka hashtable.
A list is a sequence of values.
An array is essentially the same as a list, but limited to basic datatypes. My impression is that they only exists for performance reasons, don't think I've ever used one. If performance is that critical to you, you probably don't want to use python in the first place.
A tuple is a fixed-length sequence of values (whereas lists and arrays can grow).
Lets take them one by one.
Lists:
List is a very naive kind of data structure similar to arrays in other languages in terms of the way we write them like:
['a','b','c']
This is a list in python , but seems very similar to array structure.
However there is a very large difference in the way lists are used in python and the usual arrays.
Lists are heterogenous in nature. This means that we can store any kind of data simultaneously inside it like:
ls = [1,2,'a','g',True]
As you can see, we have various kinds of data within a list and is a valid list.
However, one important thing about them is that we can access the list items using zero based indices. So we can write:
print ls[0],ls[3]
output: 1 g
Dictionary:
This datastructure is similar to a hash map data structure. It contains a (key,Value) pair. An empty dictionary looks like:
dc = {}
Now, to store a key,value pair, e.g., ('potato',3),(tomato,5), we can do as:
dc['potato'] = 3
dc['tomato'] = 5
and we saved the data in the dictionary dc.
The important thing is that we can even store another data structure element like a list within a dictionary like:
dc['list1'] = ls , where ls is the list defined above.
This shows the power of using dictionary.
In your case, you have difined a dictionary like this:
data = {'type':[],'id':[]}
This means that your dictionary will consist of only two keys and each key corresponds to a list, which are empty for now.
Talking a bit about your script, the expression :
current_row_string[-8:]
doesn't make a sense. The index should have been -6 instead of -8 that would give you the id part of the current row.
This part is the id and should have been stored in a variable say :
id = current_row_string[-6:]
Further action can be performed as seen the answer given by Roland.

How can I add to Python dictionary value using string keys

I want to add string dictionary keys like this:
x = "%s-%s-%s %s:%s:00"%(dt.year,dt.month,dt.day,dt.hour,dt.minute)
dict[x] +=a1
But it gives me an error like this:
KeyError: '2015-11-26 8:47:00'
If I try print type(x) it prints str
But if i try this:
dict = {}
x = "abc"
dict[x] = 1
print dict
it print to this:
{'abc': 1}
I don't understand what is the difference.
First error is that you named your dictionary dict. That name's
already being used; it's the name of the dictionary type. Overwriting an
existing name like this is called "shadowing". Don't do it, it will mess
you up.
You're using +=. This implies that there's already a value associated
with the key, which can be incremented. If that key isn't in the dict
yet, you get a KeyError.
You probably want to set a default value of zero. This can be done in
various ways. The simplest is:
d[x] = d.get(x, 0) + a1
Also see the collections standard library, which has a defaultdict
type.

Updating dictionary - Python

total=0
line=input()
line = line.upper()
names = {}
(tag,text) = parseLine(line) #initialize
while tag !="</PLAY>": #test
if tag =='<SPEAKER>':
if text not in names:
names.update({text})
I seem to get this far and then draw a blank.. This is what I'm trying to figure out. When I run it, I get:
ValueError: dictionary update sequence element #0 has length 8; 2 is required
Make an empty dictionary
Which I did.
(its keys will be the names of speakers and its values will be how many times s/he spoke)
Within the if statement that checks whether a tag is <SPEAKER>
If the speaker is not in the dictionary, add him to the dictionary with a value of 1
I'm pretty sure I did this right.
If he already is in the dictionary, increment his value
I'm not sure.
You are close, the big issue is on this line:
names.update({text})
You are trying to make a dictionary entry from a string using {text}, python is trying to be helpful and convert the iterable inside the curly brackets into a dictionary entry. Except the string is too long, 8 characters instead of two.
To add a new entry do this instead:
names.update({text:1})
This will set the initial value.
Now, it seems like this is homework, but you've put in a bit of effort already, so while I won't answer the question I'll give you some broad pointers.
Next step is checking if a value already exists in the dictionary. Python dictionaries have a get method that will retrieve a value from the dictionary based on the key. For example:
> names = {'romeo',1}
> print names.get('romeo')
1
But will return None if the key doesn't exist:
> names = {'romeo',1}
> print names.get('juliet')
None
But this takes an optional argument, that returns a different default value
> names = {'romeo',2}
> print names.get('juliet',1)
1
Also note that your loop as it stands will never end, as you only set tag once:
(tag,text) = parseLine(line) #initialize
while tag !="</PLAY>": #test
# you need to set tag in here
# and have an escape clause if you run out of file
The rest is left as an exercise for the reader...

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