Python cycle in lambda

Циклы и рекурсия в lambda-функциях Python

Таким образом, казалось бы, полностью исключается возможность ветвления или создания циклов внутри lambda. Вместе с тем, отчаиваться я не стал. Первым шагом к созданию цикла стал рекурсионный алгоритм:

func1 = lambda a,b=0: print(b,a) or func1(a,b+1) if b30 else None

Следует обратить внимание, что внутри lambda-функций не поддерживаются операции присваивания(=) и символы двоеточия (.
В данном алгоритме, как видите, не создается локальная переменная-счетчик. Она определяется переменной b со стандартным значением 0.
К счастью, столь ужасный алгоритм рекурсии мне не пригодился и на смену ему пришел алгоритм №2, на основе генераторов:

func2 = lambda a: [print(b,a) for b in range(30)] and None

Как видите, данный алгоритм легко использовать, слегка перестраивая его для получения определенных значений. Если планируется использовать внутри цикла только функции и методы, возвращающие None, следует поставить после выражения-генератора «and None». В этом случае функция не будет возвращать список\кортеж из NoneType объектов, а просто выполнит нужные действия нужное количество раз.
Лично мне эта функция пригодилась для стандартных «массовых методов упаковки и конфигурирования» кнопок и прочих объектов в Tkinter:

defconfigof objs,**kw: [obj.config(kw) for obj in objs] and None defpacking objs,**kw: [obj.pack(kw) for obj in objs] and None

buts = [but1,but2,but3,but4,but5] #массив объектов-кнопок для конфигурирования и упаковки defconfigof(buts,relief='groove',bg='green',fg='yellow') defpacking(buts,side='top',padx=3,pady=3,anchor='nw',fill=X)

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Python lambda for loop

In this article, we have explored how to use Lambda in Python to replicate the behavior of for loop and how to use for loop within lambda.

Table of contents:

Let us understand how lambda can be used in this case in Python.

Use for loop within lambda

Lambda can include only one line expression and for loop has multiple lines (at least 3) in Python so at first, it may seem to be difficult to use it in Lambda.

The trick is to write for loop in one line expression which can be done as follows:

As this is a one line expression, we can directly include this in a lambda. Following Python code traverses through a list of integers using lambda which in turn uses for loop:

list = [91, 12, 63, 5] f = lambda listx: [print(x) for x in listx] f(list) 

The output is as expected.

Replace for loop with lambda

Moving forward, you may want to replace a for loop which a corresponding lambda (without using for loop). This is also possible either using the lambda recursively or using reduce built-in function in Python.

Following Python code uses a lambda along with reduce function to recursively traverse through a list just like a for loop:

from functools import reduce list = [91, 12, 63, 5] f = lambda x, y: print(y) if x is None else print(x, y) a = reduce(f, list) 

Another alternative is to use the lambda recursively. This is demonstrated in the following Python code:

list = [59, 47, 7, 17, 45] f = lambda l: print(l[0]) if len(l) == 1 else print(l[0]) + f(l[1:]) sum = f(list) 

Go through this Python code to get the sum of all elements of a list using lambda just like for loop:

list = [59, 47, 7, 17, 45] f = lambda l: l[0] if len(l) == 1 else l[0] + f(l[1:]) sum = f(list) print(sum) 

Concluding Note

With this article at OpenGenus, you must have the complete idea of how to use for loop with lambda and even replace for loop with lambda.

Lambda is a powerful feature in Python. Use it wisely.

Ned Nedialkov

Professor of Computing at McMaster University | Ph.D in Computer Science from University of Toronto in 1999 | Native of Bulgaria | Research in Scientific Computing

OpenGenus Tech Review Team

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In this article, we have explored the reason uint8_t cannot be printed with std::cout and multiple fixes that can be used to print uint8_t values.

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Remove End of Line (EOL) whitespace in Files in UNIX/ Ubuntu

In this article, we have demonstrated how to Remove End of Line (EOL) whitespace in Files in UNIX/ Ubuntu using grep and sed commands.

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Python For Loop Inside Lambda

Problem: Given an iterable, how to print it using a for loop within a lambda function.

Overview

In this tutorial, we will learn why it is not a smart idea to use for loop inside Lambda in Python. This is a frequently asked question by programmers and newbies. Hence, it needs to be addressed as soon as possible.

Let us have a look at an example/question that most newbie programmers come across while dealing with the lambda function and for loops in Python. The following question is a classic example of the confusion that you might come across while using the lambda method along with a for loop.

Let’s go ahead an execute this code in our console to find out what happens!

y = "hello and welcome" x = y[0:] x = lambda x: (for i in x : print i)

x_list = list(lambda i: for i in x) ^ SyntaxError: invalid syntax

Reason: Since a for loop is a statement , it shouldn’t be included inside a lambda expression .

✨Solution 1: Using a List Comprehension

Using a lambda function with a for loop is certainly not the way to approach problems like these. Instead, you can simply use a list comprehension to iterate over the given string/list/collection and print it accordingly, as shown in the solution below.

y = "hello and welcome" x = y[0:].split(" ") res = [print(i) for i in x]

? A quick recap of list comprehensions:

List comprehension is a compact way of creating lists. The simple formula is [expression + context] .
Expression : What to do with each list element?
Context : What elements to select? The context consists of an arbitrary number of for and if statements.
Example : [x for x in range(3)] creates the list [0, 1, 2] .

✨Solution 2: Using list + map + lambda

Another workaround to our problem is to use the map() method along with a lambda function and then typecast the output to a list .

The map() function transforms one or more iterables into a new one by applying a “transformator function” to the i-th elements of each iterable. The arguments are the transformator function object and one or more iterables. If you pass n iterables as arguments, the transformator function must be an n-ary function taking n input arguments. The return value is an iterable map object of transformed, and possibly aggregated, elements.

Now, let’s have a look at the code/solution:

y = "hello and welcome" x = y[0:].split(" ") res = list(map(lambda x: print(x), x))

✨Solution 3: Using write method on sys.stdout along with join Method

A simple solution to solve the problem is to use the write method on the sys.stdout and then use the join method on the result to display the output. Let’s have a look at the following code to understand the approach:

import sys y = "hello and welcome" x = y[0:].split(" ") res = lambda x: sys.stdout.write("\n".join(x) + "\n") res(x)

Since conditionals are statements and so is print (in Python 2.x), they will not work inside the lambda expression. Thus, using the write method upon the sys.stdout module can help us to bypass this issue.

In case you are wondering about the difference between print and sys.stdout.write → refer to this link.

Note: The string.join(iterable) method concatenates all the string elements in the iterable (such as a list, string, or tuple) and returns the result as a new string. The string on which you call it is the delimiter string—and it separates the individual elements. For example, ‘-‘.join([‘hello’, ‘world’]) returns the joined string ‘hello-world’ .

?Complicated Example

The solutions to the example given above were straightforward. Now let us have a look at a slightly complicated scenario.

Problem: Given a list of numbers, store the even numbers from the list into another list and multiply each number by 5 to display the output.

The implementation of the above problem using a for loop is quite simple as shown below.

x = [2, 3, 4, 5, 6] y = [] for v in x: if v % 2: y += [v * 5] print(y)

But, how will you solve the above problem in a single line?

The solution has already been discussed above. There are two ways of doing this:

?️Method 1: Using a List Comprehension

x = [2, 3, 4, 5, 6] y = [v * 5 for v in x if v % 2] print(y) # [15, 25]

?️Method 2: Using a combination of list+map+lambda+filter

x = [2, 3, 4, 5, 6] y = list(map(lambda v: v * 5, filter(lambda u: u % 2, x))) print(y) # [15, 25]

The purpose of the above example was to ensure that you are well-versed with problems that involve iterations in a single line to generate the output.

?Tidbit: When To Use Lambda Functions?

Lambda functions are utilized when you need a function for a brief timeframe. It is also used when you have to pass a function as an argument to higher-order functions (functions that accept different functions as their arguments). Let’s have a look at a couple of examples.

lst = [1, 2, 3, 4, 5] # Adding 10 to every element in list using lambda lst2 = list(map(lambda i: i + 10, lst)) # Printing the list print("Modified List: ", lst2)

Modified List: [11, 12, 13, 14, 15]

Example 2:

# Using lambda inside another function def foo(no): return lambda i: i + no sum = foo(10) print("Sum:", sum(5))

To learn more about the lambda function and its usage in Python, please have a look at the following tutorials:
➥ Lambda Functions in Python: A Simple Introduction
➥ Lambda Calculus in Python

Conclusion

Thus, the above discussion shows it is never really a great idea to use a for loop within the lambda expressions. Instead, you can use the workarounds that have been proposed above.

To keep learning, please subscribe and stay tuned for more interesting discussions and tutorials.

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