>>> l = [0.01, 0.1, 0.4, 0.034, 0.6, 0.7, 0.9, 1] >>> [0.4 if (0. < f < 0.5) else 0.7 for f in l] [0.4, 0.4, 0.4, 0.4, 0.7, 0.7, 0.7, 0.7] solved Change the numbers of a list to another number python
Linux/macos: virtualenv -p python3 env source env/bin/activate pip install -r requirements.txt windows: virtualenv -p python3 env env\scripts\activate pip install -r requirements.txt if you need to specify the full path to python you can use something like virtualenv -p C:\Program Files\Python36\python.exe myvirtualenv solved How to create a python virtual environment from the command line? [duplicate]
If you cannot use the len() function you could write a function like num_characters below that uses a for loop to iterate over the characters in the passed in string and increments and subsequently returns a variable total based on the total amount of characters. I think that is what you mean by an accumulator … Read more
You have a list called items inside a module named items, when you call additem you are going to be checking if item in the module items not the list items, so you will get another error. Either use from items import items or change the name of your items list. Your error is most … Read more
Here is a method to do what you are trying to achieve. unsorted = [[1, 0], [2, 0], [3, 0], [4, 2], [5, 2], [6, 5], [7, 6], [8,0]] sortList = [] sortDict = {} for x in unsorted: if x[1] != 0: if x[1] in sortDict: sortDict[x[1]].append(x[0]) else: sortDict[x[1]] = [x[0]] for x in … Read more
I’m not sure what you expect this to do: “,”.join(u”%s”, (v,) for v in values) but the error message gives you some clue on what you’re doing wrong: SyntaxError: Generator expression must be parenthesized if not sole argument the “Generator expression” part here being (v,) for v in values. What you want is obviously something … Read more
Assuming the existing values don’t matter this would work def fixList(inputList, splitChar=”X”): outputList = inputList[:] x = None for i in xrange(len(outputList)): if outputList[i] == splitChar: outputList[i] = x = 0 elif x is None: continue else: outputList[i] = x x += 1 return outputList eg >>> a = [‘X’,1,2,3,4,5,6,7,8,9,’X’,11,12,13,14,15,16,17,18,19,20] >>> fixList(a) [0, 1, 2, … Read more
A variable which is defined inside a function is local to that function. It is accessible from the point at which it is defined until the end of the function and exists for as long as the function is executing. you can pass the values as arguments to other functions def naming(in_name): # function to … Read more
There’s nothing that makes recursion inherently more efficient. Moreover, the same algorithm implemented with recursion is likely to be less efficient due to function call overhead. Worse than that, and this is exactly your case, using recursions of large depth is likely to cause stack overflow (no pun intended). Unless tail-recursion optimization is in use, … Read more
Create your html file as a string in python. This string can be modified in response to sensor input. Write it to a file on your system where the html is source file exists; rewrite each time the sensor inputs change. Set up a timer in Javascript, e.g. setTimeout() to refresh the page often. 2 … Read more
Something like this should work… df = pd.DataFrame({‘date’: [‘2017-01-01 01:01:01’, ‘2017-01-02 01:01:01’, ‘2017-01-03 01:01:01’, ‘2017-01-30 01:01:01’, ‘2017-01-31 01:01:01’], ‘value’: [99,98,97,95,94]}) df[‘date’] = pd.to_datetime(df[‘date’]) def get_list(row): subset = df[(row[‘date’] – df[‘date’] <= pd.to_timedelta(‘5 days’)) & (row[‘date’] – df[‘date’] >= pd.to_timedelta(‘0 days’))] return str(subset[‘value’].tolist()) df[‘list’] = df.apply(get_list, axis=1) Output: date value list 0 2017-01-01 01:01:01 99 [99] … Read more
I do not fully understand what you wanted to achieve but i think is one of the following 2: 1.return a continuous range of value between minimum and maximum. In this case you can use range and do not pass Y as input param. def Z(minimum,maximum): Y = range(minimum, maximum) return Y print Z(2,4) 2.you … Read more
There is an array array1=[20,10,4,3,8,9,30]. There result array should have the immediate next number of the main array. i.e array2=[30,20,8,4,9,30] [closed] solved There is an array array1=[20,10,4,3,8,9,30]. There result array should have the immediate next number of the main array. i.e array2=[30,20,8,4,9,30] [closed]
You can use ast.literal_eval to safely evaluate strings containing Python literals. from ast import literal_eval a=”[“Hello”, “World!”, 2]” b = literal_eval(a) # [“Hello”, “World!”, 2] Note that the string can only be compromised of: strings, bytes, numbers, tuples, lists, dicts, sets, booleans, and None (taken from the documentation here) solved how to convert string consisted … Read more
>>> from itertools import groupby >>> from operator import itemgetter >>> list1 [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>> list2 [0, 0, 3, 4, 5, 0, 0, 8, 9, 0] >>> [[e[0] for e in v] for k,v in groupby(((a ,b, a==b) for a,b in zip(list1, list2)), itemgetter(2)) if k] [[3, … Read more