You need to set$final[$id] to an array before adding elements to it. Intiialize it with either $final[$id] = array(); $final[$id][0] = 3; $final[$id][‘link’] = “https://stackoverflow.com/”.$row[‘permalink’]; $final[$id][‘title’] = $row[‘title’]; or $final[$id] = array(0 => 3); $final[$id][‘link’] = “https://stackoverflow.com/”.$row[‘permalink’]; $final[$id][‘title’] = $row[‘title’];
Try a list comprehension: l = [x * 2 for x in l] This goes through l, multiplying each element by two. Of course, there’s more than one way to do it. If you’re into lambda functions and map, you can even do l = map(lambda x: x * 2, l) to apply the function … Read more
The title of your question is a bit misleading. Fortran doesn’t allow you to pass a scalar to an array. But what it DOES allow is passing a single element of an array to a routine’s array dummy argument – this is called “sequence association” in Fortran. As IanH and others have said, the following … Read more
Here’s an example which issues the same warning: import numpy as np np.seterr(all=”warn”) A = np.array([10]) a=A[-1] a**a yields RuntimeWarning: overflow encountered in long_scalars In the example above it happens because a is of dtype int32, and the maximim value storable in an int32 is 2**31-1. Since 10**10 > 2**32-1, the exponentiation results in a … Read more
Non-numpy functions like math.abs() or math.log10() don’t play nicely with numpy arrays. Just replace the line raising an error with: m = np.log10(np.abs(x)) Apart from that the np.polyfit() call will not work because it is missing a parameter (and you are not assigning the result for further use anyway).
In PSv3+, with its unified handling of scalars and collections, any object – even $null – should have a .Count property (and, with the exception of $null, should support indexing with [0]). Any occurrence of an object not supporting the above should be considered a bug; for instance: Using this intrinsic (engine-supplied) .Count property unexpectedly … Read more
Since PHP 5.6, you can declare an array constant with const: <?php const DEFAULT_ROLES = array(‘guy’, ‘development team’); The short syntax works too, as you’d expect: <?php const DEFAULT_ROLES = [‘guy’, ‘development team’]; If you have PHP 7, you can finally use define(), just as you had first tried: <?php define(‘DEFAULT_ROLES’, array(‘guy’, ‘development team’));
The error message says that if you’re passing scalar values, you have to pass an index. So you can either not use scalar values for the columns — e.g. use a list: >>> df = pd.DataFrame({‘A’: [a], ‘B’: [b]}) >>> df A B 0 2 3 or use scalar values and pass an index: >>> … Read more