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How assignment works with Python list slice?
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4 Answers
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Active
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Apply a function to every row of a matrix or a data frame
...:6, nrow=3, byrow=TRUE)
R> M
[,1] [,2]
[1,] 1 2
[2,] 3 4
[3,] 5 6
R> apply(M, 1, function(x) 2*x[1]+x[2])
[1] 4 10 16
R>
This takes a matrix and applies a (silly) function to each row. You pass extra arguments to the function as fourth, fifth, ... arguments to app...
Version of SQLite used in Android?
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482
UPDATE OCT 2016: Here is a link to the updated official docs which includes the main points in...
NameError: global name 'xrange' is not defined in Python 3
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answered Jun 19 '13 at 13:14
Martijn Pieters♦Martijn Pieters
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Searching for UUIDs in text with regex
...y I'm relying on the assumption that all UUIDs will follow a patttern of 8-4-4-4-12 hexadecimal digits.
16 Answers
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No secret option provided to Rack::Session::Cookie warning?
...n be safely ignored by Rails users.
(https://github.com/rack/rack/issues/485#issuecomment-11956708, emphasis added)
Confirmation on the rails bug discussion: https://github.com/rails/rails/issues/7372#issuecomment-11981397
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Transpose list of lists
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How about
map(list, zip(*l))
--> [[1, 4, 7], [2, 5, 8], [3, 6, 9]]
For python 3.x users can use
list(map(list, zip(*l)))
Explanation:
There are two things we need to know to understand what's going on:
The signature of zip: zip(*iterables) This means zip ...
Creating an instance of class
What's the difference between lines 1 , 2 , 3 , 4?
3 Answers
3
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How to add a new row to an empty numpy array
...proper dimensionality.
>>> arr
array([], shape=(0, 3), dtype=int64)
Then be sure to append along axis 0:
arr = np.append(arr, np.array([[1,2,3]]), axis=0)
arr = np.append(arr, np.array([[4,5,6]]), axis=0)
But, @jonrsharpe is right. In fact, if you're going to be appending in a loop, ...
How do I make a list of data frames?
...nt you're working in:
d1 <- data.frame(y1 <- c(1, 2, 3), y2 <- c(4, 5, 6))
y1
# [1] 1 2 3
y2
# [1] 4 5 6
This won't have the seemingly desired effect of creating column names in the data frame:
d1
# y1....c.1..2..3. y2....c.4..5..6.
# 1 1 4
# 2 ...