You may simply map the split, filter if it makes a size 2 tuple, and zip it to get the values separately: In[9]: l=[‘Name’, ‘AMAZON.COM – TOT RETURN IND’, ‘AMAZON.COM INC – NET CASH FLOW – FINANCING’, ‘AMAZON.COM INC – NET CASH FLOW – INVESTING’, ‘AMAZON.COM INC – NET CASH FLOW-OPERATING ACTIVS’, ‘AMAZON.COM INC – … Read more
You could do something like this: string = ” for t in range(1,8): if t % 2 == 0: # if t is even multiply t by 3 and add t to the string string += str(t * 3) + ‘ ‘ else: # if t is odd simply add t to the string string … Read more
Fixing your indentation: play = input(‘| Please enter h or s (h = Hit, s = Stand:) ‘) while play == ‘s’ and totalplayer <= 15: print(‘Cannot stand on value less then 15’) while loops keep executing continuously while the loop conditions are true. There is no code inside your loop body that would change … Read more
For start i suggest for you to start with Tkinter library (built in library). This is a simple program using Tkinter gui. import Tkinter class simpleapp_tk(Tkinter.Tk): def __init__(self,parent): Tkinter.Tk.__init__(self,parent) self.parent = parent self.initialize() def initialize(self): self.grid() self.entryVariable = Tkinter.StringVar() self.entry = Tkinter.Entry(self,textvariable=self.entryVariable) self.entry.grid(column=0,row=0,sticky=’EW’) self.entry.bind(“<Return>”, self.OnPressEnter) self.entryVariable.set(u”Enter text here.”) button = Tkinter.Button(self,text=u”Click me !”, command=self.OnButtonClick) button.grid(column=1,row=0) … Read more
All dependent lines must be indented: def MakeOneFeedScript(): z = 1 CFeedList = [str(“null”)]*3 #Liste mit Länge der Feeds Erstellen # Listen Für 1. 2. 3. Post, usw, IFeedList = [] #Liste für Posttime des 1. posts pro feed IIFeedList = [] #Liste für Posttime des 2. posts pro feed IIIFeedList = [] #Liste für … Read more
>>> from collections import OrderedDict >>> sorted_dict = OrderedDict() >>> dct = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0} >>> dct.values().sort() >>> for x in sorted(dct.values(), reverse=True): … keys = [idx for idx in dct if dct[idx]==x] … for key in keys: … sorted_dict[key] = x … >>> >>> sorted_dict OrderedDict([(3, 4), … Read more
I’m not sure what you’re trying to do because I don’t understand your question that well. Unfortunately, the code runs insanely slow. (It takes about 75 minutes to complete.) Mabey you should look for a different solution. I copied it out for the image and got: x = [1.0, 2.0, 3.0] y = [1.0, 3.0, … Read more
The formula is pretty basic, it say: The function ‘f’ for provided argument ‘S’ returns value ‘K’ if value ‘S’ is less or equal then the value ‘K’, if the value ‘S’ is greater then the value ‘K’ AND lower then the value ‘2*K’ – return value ‘2*K-S’, otherwise return 0. Python: def A(S,K): result … Read more
It’s a very subjective question, but Unity sounds like what you’re looking for. It has support for Boo as a language (although it’s not at the forefront of Unity usage). Boo isn’t Python, but is similar enough for you to carry on your knowledge of Python with you, and Unity is a solid engine for … Read more
if you are expecting “.google.” then none of these options will work but this might: word=’www.wgoowglew.cowm’ print ‘.’ + word.split(‘.’)[1].replace(‘w’, ”) + ‘.’ solved Remove lowercase letters from a string in python
Yes, a generator can’t be equal to a string. When your call your function: get_objects(“anything”), it returns a generator object you can iterate over to get the values. If you want to check if the ith element returned is equal to something, do: for i, elem in enumerate(get_objects(“bla bla”)): if i == 3: return self.assertEqual(elem, … Read more
Read a specific item in index i: item = L[index] Read a sequence of items from index ‘start’ to index ‘stop’: seq = L[start:stop] Iterate over a list: for item in L: print item If you need both the index and the item, use the enumerate function: for index, item in enumerate(L): print index, item … Read more
For continue, it essentially skips to the end of the loop and starts the next loop, so following through your loop: Look at the current letter If the current letter is a vowel, then continue. (This skips to the next letter in the loop, so the line print(letter) will not be run). If the current … Read more
Use a total variable and add the generated number to the total import random x = 0 total = 0 while x<10: number = random.randrange(1,10) print(number) total += number x=x+1 print(total) 0 solved i’ve generated random numbers. Now i want to add all the numbers that are generated. What is the code to add them … Read more
Else block cannot be kept empty. It must at-least contain ‘pass’. The line ‘print(error)’ must be indented inside the else block. Also your ‘Cmdbar()’ in the elif must be indented. 0 solved “Statement expected” on else but there’s nothing wrong with the code?