You can do like this: a, b, c, d, e = input(“Insert the 5 values: “).split() print(f”a: {a}\nb: {b}\nc: {c}\nd: {d}\ne: {e}”) Input: 1 2 3 4 5 Output: a: 1 b: 2 c: 3 d: 4 e: 5 1 solved How can I input multiple values in one line? [duplicate]
Just disable html extension and enable django extension in vscode. I had that kind of problem. 0 solved Blocks are not highlighting in django vs code [closed]
Better to use built-in function max in module builtins: >>> data = [73284, 8784.3, 9480938.2, 984958.3, 24131, 45789, 734987, 23545.3, 894859.2, 842758.3] >>> max(data) 9480938.2 Info Page: max(iterable, *[, default=obj, key=func]) -> value max(arg1, arg2, *args, *[, key=func]) -> value With a single iterable argument, return its biggest item. The default keyword-only argument specifies an … Read more
You can use DataFrame.append and then group by the column you want to use as an index. Let’s say that total_df is the table to which you want to add new rows, new_df is the table containing those new rows and date is the column to be used as index. Then you can use: total_df.append(new_df).groupby(by=’date’).sum() … Read more
This is the syntax of a dictionary in Python, a data structure which maps an index to another object. a = tf.placeholder(tf.float32) {a: [1,3]} In this case, a is a Tensor from Tensorflow and [1,3] are the values it will assume when the graph is executed. IMO, you should invest your time into acquiring a … Read more
For anyone who has the same problem in the future, the solution is to remove the extra line. writer=csv.DictWriter(f,fieldnames=fieldnames, lineterminator=”\n”) Although it can be read with the extra line spacing by changing This: namesList = [x[0] for x in people] To this: namesList = [x[0:1] for x in people] (in my example) blank results are … Read more
You can do that: a = [“first answer”, “second answer”, “last answer”] answer = raw_input(“\n”.join([chr(x + ord(‘A’)) + “) ” + a[x] for x in xrange(len(a))])) Edit (for Python 3): a = [“first answer”, “second answer”, “last answer”] answer = input(“\n”.join([chr(x + ord(‘A’)) + “) ” + a[x] for x in range(len(a))])) Explanation: for x … Read more
May be this function can help: def checkChange(FileName, oldContent): fSrc = open(FileName, ‘r’) actualContent = fSrc.read() fSrc.close() if actualContent != oldContent: print(1) return actualContent solved Python Check txt variable loop [closed]
We have some list of lists a. We want to get counts for each of the sublists, and the total of those counts. Note that the total counts will just be the sum of the counts of the sublists. from collections import Counter a = [[‘das’,’sadad’,’asdas’,’das’],[‘das’,’sadad’,’da’],[‘aaa’,’8.9′]] def count(list_of_lists): counts = [Counter(sublist) for sublist in list_of_lists] … Read more
Update: for multiple dictionaries Iterate over your dictionaries and for every one of them, check if the value of ‘mesic’ is in [6,7,8] and if so, get the corresponding dictionary values: d1 = {‘stat’: u’AS’, ‘vyska’: 3.72, ‘stanice’: u’AQW00061705′, ‘mesic’: 8, ‘teplotaC’: 26.88} d2 = {‘stat’: u’AS’, ‘vyska’: 3.72, ‘stanice’: u’AQW00061705′, ‘mesic’: 1, ‘teplotaC’: 26.88} … Read more
How to perform self join with same row of previous group(month) to bring in additional columns with different expressions in Pyspark solved How to perform self join with same row of previous group(month) to bring in additional columns with different expressions in Pyspark
import json import requests response = requests.get(…) json_data = json.loads(response.text) solved how to parse the response in python?
There is not (much) difference between def chain_for(*a): and def __init__(self, *a):. Hence, a very crude way to implement this can be: class chain_for: def __init__(self, *lists): self.lists = iter(lists) self.c = iter(next(self.lists)) def __iter__(self): while True: try: yield next(self.c) except StopIteration: try: self.c = iter(next(self.lists)) except StopIteration: break yield next(self.c) chain = chain_for([1, 2], … Read more
int(x) yields your error, since you can not convert “A” into an integer Correct would be: a=”ABCD” b=map(lambda x:x,a) print(list(b)) As mentioned in the comments, the following gives the same result: print(list(a)) You should probably check out some more lambda tutorials first: http://www.secnetix.de/olli/Python/lambda_functions.hawk solved Using lambda function in python
There must be something wrong with the quiz, the code as written in the question is declaring that MyInt is a subclass of int (or equivalently: that int is the superclass of MyInt), no the other way around. 1 solved Superclass inherits from a subclass. Coursera. Are they crazy? [closed]