my_dict = {key1:{value1:value2}, key2:{value3:value4}, key3{value5:value6}, key4:{value7:value8}…} result = [] for key1 in my_dict: if sum(my_dict[key1].keys() + my_dict[key1].values())==3: result[key1] = my_dict[key1].keys()[0] Hope this helps solved Python – selectively add up values while iterating through dictionary of dictionaries [closed]
So I found out after reading through the documentation that I typed in the command wrong. Here is the new command I used pyinstaller -w –icon=AppIcon.icns VideoDownloader.py. Works like a charm. 1 solved My mac app I made with pyinstaller opens and then closes immediately but then reopens 30 seconds later
First, you start your question with expenses but ends with incomes and the code also has incomes in the parameters so I’ll go with the income. Second, the error says the answer. “expense_types(income_types)” is a list and “expenses(incomes)” is a dictionary. You’re trying to find a list (not hashable) in a dictionary. So to make … Read more
I’ve added comments to the answer to explain the flow, please add more details in your question if this is not clear enough 🙂 # Open file for reading file_house_price = open(“house_price.txt”, “r”) # read all lines from the file into a variable data_house_price = file_house_price.readlines() # as we’re done with the file, we can … Read more
This will add a numbered (name, job) pair for each name: dict = {} i = 0 # or 1, if you prefer. but it’s not proper to start a list from 1 when programming. while i < len(names): dict[i] = (names[i], Jobs[i]) i = i+1 as a for loop: for i in id_: dict[i] … Read more
Right now, the else clause fires if there is a at least one key that does not match vehicle. Fix it like this: def add_to_dict(product_type, brand, product_dict): if product_type in product_dict: product_dict[product_type].append(brand) else: product_dict[product_type] = [brand] product_dict = {} add_to_dict(“car”, “audi”, product_dict) add_to_dict(“car”, “mercedes”, product_dict) add_to_dict(“phone”, “apple”, product_dict) print(product_dict) # Output: # {“car”: [“audi”,”mercedes”], “phone”: … Read more
Try this one-liner: reduce(lambda acc, cur: acc + ([] if cur[1] == ‘you’ else [next((i[0] – cur[0] – 1 for i in list(enumerate(my_list))[cur[0]+1:] if i[1] == ‘hello’), len(my_list) – cur[0] – 1)]), enumerate(my_list), []) It will give an array where the nth element is the number of ‘you’s following the nth occurrence of ‘hello’. e.g. … Read more
If I am interpreting your question correctly, you simply need to devise a very simple list comprehension program. Getting the square root of everything between zero and n is not necessary. Did you mean square? Here’s my solution: def nbDig(n,d): int_list = str([str(x*x) for x in range(0,n+1)]).count(str(d)) return int_list I can test it like so: … Read more
You don’t need this void setup() Linux is take a care of this. Here is Python code: import RPi.GPIO as GPIO import time #GPIO Mode (BOARD / BCM) GPIO.setmode(GPIO.BCM) #set GPIO Pins GPIO_TRIGGER = 18 GPIO_ECHO = 24 #set GPIO direction (IN / OUT) GPIO.setup(GPIO_TRIGGER, GPIO.OUT) GPIO.setup(GPIO_ECHO, GPIO.IN) def distance(): # set Trigger to HIGH … Read more
If I read your problem correctly, you need to provide a method to compensate for the broken coin producing a relatively fair amount of results. With that instead of calling the broken_coin() function every time directly, one could call a function that calls the function and every other time returns the reverse result to the … Read more
x = [‘cat’, ‘dog’, ‘chicken’] y = [word[0] + ‘_’*(len(word)-1) for word in x] print(y) # [‘c__’, ‘d__’, ‘c______’] solved how to show only the first letter of an element of a string [closed]
Three separate tasks: call svn properly to create the log parse the log Write the parsed values somewhere 1. import subprocess as sp svn_url = “svn://repo-path.com/project” revisions = [12345, 12346] revision_clargs = [“-r%i” % revision for revision in revisions] popen = sp.Popen([“svn”, “log”, “-v”] + revision_clargs + [svn_url],stdout=sp.PIPE,stderr=sp.PIPE) out,err = popen.communicate() 2. input_ = “”” … Read more
color_img = imread(‘0k4Kh.jpg’); img = rgb2gray(color_img); [x, y] = size(img); for i = 1:x if length(find(img(i, :))) ~= 0 lastmarginalrow = i-1; break; end end for ii = y:-1:1 if length(find(img(:, ii))) ~= 0 lastmarginalcol = ii-1; break; end end figure; fig = imshow(color_img); h = impoly(gca, [0,x; lastmarginalcol,x; lastmarginalcol,lastmarginalrow; 0,lastmarginalrow]); api = iptgetapi(h); api.setColor(‘red’); … Read more
Keeping as much as possible your code structure, this should be the solution you are looking for. It’s meant to be easy to read and understand. However it is not the best solution, for that we need to know what selection looks like, or even better what your input file looks like. def main(): with … Read more
If you want this to work for any arbitrary key(s) you can use a defaultdict of OrderedDicts.. from collections import defaultdict, OrderedDict result_dict = defaultdict(OrderedDict) data = [(‘Han Decane’,’12333′),(‘Can Decane’,’12333′),(‘AlRight’,’10110′)] for (v,k) in data: result_dict[k][v]=True >>> list(result_dict[‘12333’].keys()) [‘Han Decane’, ‘Can Decane’] And if you want all the results that had multiple values >>> [k for … Read more