- Python style guide recommends using list comprehensions instead of map/reduce
- String formatting using percent operator is obsolete, consider using format() method
the code you need is this simple one-liner
output = [" this string contains {} and {}".format(x, y) for (x, y) in matrix]
Hello Pythonistas!
I've been on a Python journey recently, and I've found myself fascinated by the power and flexibility of Lambda functions. These anonymous functions have not only made my code more efficient and concise, but they've also opened up a new way of thinking about data manipulation when used with Python's built-in functions like Map, Filter, and Reduce.
Lambda functions are incredibly versatile. They can take any number of arguments, but can only have one expression. This makes them perfect for small, one-time-use functions that you don't want to give a name.
Here's a simple example of a Lambda function that squares a number:
square = lambda x: x ** 2
print(square(5)) # Output: 25
But the real power of Lambda functions comes when you use them with functions like Map, Filter, and Reduce. For instance, you can use a Lambda function with `map()` to square all numbers in a list:
numbers = [1, 2, 3, 4, 5]
squared = list(map(lambda x: x ** 2, numbers))
print(squared) # Output: [1, 4, 9, 16, 25]
You can also use a Lambda function with `filter()` to get all the even numbers from a list:
numbers = [1, 2, 3, 4, 5]
even = list(filter(lambda x: x % 2 == 0, numbers))
print(even) # Output: [2, 4]
And finally, you can use a Lambda function with `reduce()` to get the product of all numbers in a list:
from functools import reduce
numbers = [1, 2, 3, 4, 5]
product = reduce(lambda x, y: x * y, numbers)
print(product) # Output: 120
Understanding and using Lambda functions, especially in conjunction with Map, Filter, and Reduce, has significantly improved my data manipulation skills in Python. If you haven't explored Lambda functions yet, I highly recommend giving them a try!
Happy coding!
Mapping arrays in Python using map, lambda, and functional programming workflows - Stack Overflow
arrays - python lambda map over nested list - Stack Overflow
performance - Efficient way to use python's lambda, map - Stack Overflow
Python: Lambda, Map, Filter, Reduce Functions explained in 1 simple video
You're still making python2 tutorials? Why?
More on reddit.comVideos
- Python style guide recommends using list comprehensions instead of map/reduce
- String formatting using percent operator is obsolete, consider using format() method
the code you need is this simple one-liner
output = [" this string contains {} and {}".format(x, y) for (x, y) in matrix]
You have a couple of issues, these structures aren't nested deeply enough to warrant the nested loops.
You need 1 map for each level of list you wish to process, so if you want to process a list, you need a map, if you want to process a list of lists, you need 2 and so on.
In this case you most likely only want to process the top level (effectively this is because you want each list in the top level to become a sentence).
def sentence( x, y):
return " this string contains %s and %s" % (x,y)
matrix = [['a','b'],['c','d']]
output = map(lambda a: sentence(a[0],a[1]), matrix)
# Print the top level
for i in output:
print(i)
While I'm not certain of this, it sounds like the input might have the length included in the beginning and it needs to be passed through untouched. If that's the case, here's a single lambda function that will take the input and produce the desired output:
lambda i: [n**2 for n in i if n > 0] if isinstance(i, list) else i
This portion does the squaring on the non-negatives: [n**2 for n in i if n > 0]
But only if the value is a list: if isinstance(i, list)
Otherwise pass the value through: else i
That means this input [2, [1, 2, 3, -1, 2], [2, 4, -3]] returns this output [2, [1, 4, 9, 4], [4, 16]]
You have to use a lambda function within a lambda function. The result isn't that pretty, but it works:
list(map(lambda x: list(map(lambda y: y ** 2, filter(lambda z: z > 0, x))), arr[1:]))
With the given input case, this outputs:
[[1, 4, 9, 4], [4, 16]]
If you can't use list slicing, you could use filter(lambda x: isinstance(x, list), arr) rather than arr[1:].
Let's start at the innermost lambda functions and work our way outwards. To simplify things, we start by considering how to perform this operation on a single list, called x:
filter(lambda z: z > 0, x)gives us all the positive elements inx.- We then square all of these elements using
map:map(lambda y: y ** 2, ...) - This gives us
map(lambda y: y ** 2, filter(lambda z: z > 0, x)).
This gives us something that works for a single list. How do we extend it to work with a list of lists? Well, the operation is defined in the same way for each list, so use map again! (slicing off the first element because it isn't a list, and transforming the map objects into lists to match the desired output).
This finally gives us:
list(map(lambda x: list(map(lambda y: y ** 2, filter(lambda z: z > 0, x))), arr[1:]))
as specified originally.
For such large data structures numpy will work well. For this example, it's over 200x faster (see below), and a bit easier to code, basically just
add.accumulate(diff)
Comparison between numpy and direct list manipulation:
import numpy as nx
import timeit
N = 10000
diff_nx = nx.zeros(N, dtype=nx.int)
diff_py = list(diff_nx)
start = 1005
def f0():
orig = [start]
for x in diff_py:
orig.append(orig[-1] + x)
def f1():
diff_nx[0] = start
nx.add.accumulate(diff_nx)
t = timeit.Timer("f0()", "from __main__ import f0, f1, diff_nx, diff_py, nx, start")
print t.timeit(number=1000)
t = timeit.Timer("f1()", "from __main__ import f0, f1, diff_nx, diff_py, nx, start")
print t.timeit(number=1000)
gives
13.4044158459 # for list looping
0.0474112033844 # for numpy accumulate
Really, though, it seems better to reuse an established compression algorithm, like can easily be done with PyTables, rather than rolling your own like it seems that you're doing here.
Also, here, I'm suggesting that you read in the data with room for the prepended start term, rather than rebuild the list with the prepended term, of course, so you don't have to do the copy.
The following works for me:
orig = [start]
for x in diff:
orig.append(orig[-1] + x)
Using map will create an new array of the same size, filled with None. I also find a simple for loop more readable, and in this case as fast as you can get.