yes - lambda is actually "make function"; you will need to give it a name

lst1 = [lambda x:x*i for i in range(5)]

def replace_lambda(x):
    return x * x

lst2 = [replace_lambda for i in range(5)]

print lst1
print lst2

for idx, func in enumerate(lst1):
    print func(idx)

for idx, func in enumerate(lst2):
    print func(idx)

result:

[<function <lambda>>, <function <lambda>>, <function <lambda>>, <function <lambda>>, <function <lambda>>]
[<function replace_lambda>, <function replace_lambda>, <function replace_lambda>, <function replace_lambda>, <function replace_lambda>]
0
1
4
9
16
0
1
4
9
16

You can't really do it that easily and if you could a lambda function is still not designed for your use case.

Multiple replacements like that are done using a regular for loop statement, and a lambda is limited to a single expression. If you have to use a function, use a normal function – it's entirely equivalent to a lambda function except that it's not limited to a single expression.

If you really must know how to do it in a single expression, you have three choices:

1) If you use unicode strings (or Python 3), and limit your bad substrings to single characters (i.e. remove "\newline"), you can use the unicode.translate method.

bad_chars = {u"\n": u" ", u"\b": u" ", u"\f": u" ", u"\r": u" ", u"\t": u" ", u"\v": u" ", u"\x00": u" "}
bad_chars_table = dict((ord(k), v) for k, v in bad_chars.iteritems())
translator = lambda s: s.translate(bad_chars_table)
print translator(u"here\nwe\tgo")

2) Use regular expressions:

   translator = lambda s: re.sub(r'[\n\b\f\r\t\v\x00]', ' ', s)

3) You can use reduce which can be used to reduce a sequence using a binary operation, essentially repeatedly calling a function of two arguments with the current value and an element of the sequence to get the next value.

translator = lambda s: reduce(lambda x, (from, to): x.replace(from, to), bad_chars.iteritems(), s)

As you can see, the last solution is much more difficult to understand than:

def translator(s):
    for original, replacement in bad_chars.iteritems():
        s = s.replace(original, replacement)
    return s

And both solutions do the same thing. It's often better to program for the end, not for the means. For an arbitrary problem a comprehensible single-expression solution wouldn't exist at all.

To avoid any confusion, lets first extract the lambda function and see what it does. The lambda function is defined as

lambda x: x if(x != a1 and x != a2) else a1 if x == a2 else a2

A lambda statement avoids the definition of creating a named function, however any lambda function can still be defined as a normal function. This specific lambda function uses a ternary operator. This can be expanded to a regular if-else statement. This would lead to an equivalent regular function like this

def func(x, a1, a2):
    if x != a1 and x != a2:
        return x
    elif x == a2:
        return a1
    else:
        return a2

Use a generator:

return sum(x.fee for x in self.bookings.values())

Generally, generators and comprehensions improve readability, especially if you have to use a lambda to make map or filter work. Moreover, map and lambda don't play well together performance-wise (Python List Comprehension Vs. Map).

While the str.replace can only replace one substring with another, re.sub can replace a pattern.

In [55]: eSToAvoid = 'éêèÉÊÈ' 
In [58]: import re 

test cases:

In [61]: re.sub(r'[éêèÉÊÈ]', 'e', 'foobar')                                                                          
Out[61]: 'foobar'
In [62]: re.sub(r'[éêèÉÊÈ]', 'e', eSToAvoid)                                                                         
Out[62]: 'eeeeee'
In [63]: re.sub(r'[éêèÉÊÈ]', 'e', 'testingè,É  foobar  è É')                                                         
Out[63]: 'testinge,e  foobar  e e'

The string replace approach is:

In [70]: astr = 'testingè,É  foobar  è É' 
    ...: for e in eSToAvoid: 
    ...:     astr = astr.replace(e,'e') 
    ...:                                                                                                             
In [71]: astr                                                                                                        
Out[71]: 'testinge,e  foobar  e e'

the replace is applied sequentially to astr. This can't be expressed as a list comprehension (or map). A list comprehensions most naturally replaces a loop that collects its results in a list (with list.append).

There's nothing wrong with the for loop. It's actually faster:

In [72]: %%timeit 
    ...: astr = 'testingè,É  foobar  è É' 
    ...: for e in eSToAvoid: 
    ...:     astr = astr.replace(e,'e') 
    ...:  
    ...:                                                                                                             
1.37 µs ± 8.96 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [73]: timeit re.sub(r'[éêèÉÊÈ]', 'e', 'testingè,É  foobar  è É')                                                  
2.79 µs ± 15.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [77]: timeit astr.translate(str.maketrans(eSToAvoid, 'e' * len(eSToAvoid)))                                       
2.56 µs ± 14.5 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

reduce

In [93]: from functools import reduce  
In [96]: reduce(lambda s,e: s.replace(e,'e'),eSToAvoid, 'testingè,É  foobar  è É' )                                  
Out[96]: 'testinge,e  foobar  e e'
In [97]: timeit reduce(lambda s,e: s.replace(e,'e'),eSToAvoid, 'testingè,É  foobar  è É' )                           
2.11 µs ± 32.1 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

For fun you could also explore some of the idea presented here:

Cleanest way to combine reduce and map in Python

You want to 'accumulate' changes, and to do that, you need some sort of accumulator, something that hangs on to the last replace. itertools has an accumulate function, and Py 3.8 introduced a := walrus operator.

generator

In [110]: def foo(astr, es): 
     ...:     for e in es: 
     ...:         astr = astr.replace(e,'e') 
     ...:         yield astr 
     ...:                                                                                                            
In [111]: list(foo(astr, eSToAvoid))                                                                                 
Out[111]: 
['testingè,É  foobar  è É',
 'testingè,É  foobar  è É',
 'testinge,É  foobar  e É',
 'testinge,e  foobar  e e',
 'testinge,e  foobar  e e',
 'testinge,e  foobar  e e']

Or [s for s in foo(astr, eSToAvoid)] in place of the list(). This highlights that fact that a list comprehension returns a list of strings, even if the strings accumulate the changes.

I don't think you can put while-loops inside lambdas, and if you could, you shouldn't. You can get around this with map or similar, but be aware of the Zen of Python.

Lambda expressions are usually short and easy to read. Your print_limit function is more readable as it is.

I agree that using regex matching is a good solution to your problem, but I can at least address the problem with your code.

You current code is:

df['tel_no'].apply(lambda x: x.replace(i, '') for i in ['+','-','tel:'])

Python parses this (perhaps surprisingly) as:

df['tel_no'].apply(
    (
        (lambda x: x.replace(i, ''))
        for i in ['+','-','tel:'])
    )
)

That is, you have written a generator comprehension, creating a new anonymous function at each iteration of the loop. You have not created a single anonymous function with a generator comprehension inside it!

Obviously, generators are not callable, which is what caused the error.

Your attempt reflects two additional misunderstandings:

1. Comprehension syntax cannot be used outside of an actual comprehension. Perhaps you meant to write lambda x: (x.replace(i, '')) for i in ['+','-','tel:']), which would at least be one function that contains a generator comprehension.

2. String functions like str.replace do not modify the string. They return a new string. See the example below.

s1 = 'hello'
s2 = s1.replace('e', 'f')

# s1 will be unchanged
assert s1 == 'hello'

# s2 will be changed
assert s2 == 'hfllo'

To write this as a function, you would need to use def, not `lambda:

def clean_tel(x):
    for bad_string in ['+', '-', 'tel:']:
        x = x.replace(bad_string, '')
    return x

df['tel_no'].apply(clean_tel)

Or you can omit the loop and write it like this:

df['tel_no'].apply(
    lambda x: x.replace('+', '').replace('-', '').replace('tel:', '')
)

Pandas.Dataframe.replace() doesn't accept a function as the replacement value, so it simply converts it to a string.

To use a function replacement like in re.sub() you have to use Pandas.Series.str.replace()

df.Date = df.Date.str.replace(r'(\d+/\d+/ \d+)', lambda x: x.group(1).replace(" ", ""), regex=True)
print(df)

prints

    Courses    Fee Duration  Discount       Date
r1    Spark  20000    30day      1000  23/3/2022
r2  PySpark  25000   40days      2300    99/5/34
r3   Hadoop  26000      NaN      1500   7/7/2122
r4      NaT  22000     None      1200  6/12/2024