Try

lambda x: 1 if x == "C" else 0

possible duplicate of Is there a way to perform "if" in python's lambda

Example :

map(lambda x: True if x % 2 == 0 else False, range(1, 11))

result will be - [False, True, False, True, False, True, False, True, False, True]

You'll have to wrap the map around a filter around the list:

example_map = map(lambda x: x*2, filter(lambda x: x*2/6. != 1, range(5)))

>>> lst = [1,2,4,5]
>>> map(lambda x: 'lower' if x < 3 else 'higher', lst)
['lower', 'lower', 'higher', 'higher']

Aside: It's usually preferred to use a list comprehension for this

>>> ['lower' if x < 3 else 'higher' for x in lst]
['lower', 'lower', 'higher', 'higher']

map always produces one output item for each input item, it can not remove elements. Furthermore, map should not be used to mutate objects, that's not its job, and because it's lazy the results can be unexpected.

filter is designed to create an output with less elements than the input, although it's mostly useful if you already have a ready-made predicate (filtering) function.

Since you do not, you can and should use comprehensions which provide a relatively terse way to perform iteration, filtering, mapping and collection in a single construct:

wordlist = ['hello','world','Tom']
checklist = ['hello','world']

print('before')
print(wordlist)

wordlist = [word for word in wordlist if word not in checklist]
print('after')
print(wordlist)

ps: if you want to modify things in-place, use a regular loop

You are splitting b1 on single spaces, and this can lead to empty values:

>>> '88  89 '.split(' ')
['88', '', '89', '']

It is the extra empty strings here that cause int() to throw an exception:

>>> int('')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: invalid literal for int() with base 10: ''

Use str.split() with no argument instead; extra whitespace is then stripped:

>>> '88  89 '.split()
['88', '89']

You have some other problems as well in your code. Take a good look at:

def lab8():
    userinput= "Lab8.txt"
    lenoffile= len(userinput)
    print "There is", lenoffile, "lines"
    File= open (userinput, "r")
    studentscores1= File.read()
    studentlist= studentscores1.split("\n")
    return studentlist, lenoffile

Here, lenoffile is not the number of lines in the file. It is the number of characters in 'Lab8.txt'; both values happen to be 8, but add or remove some lines from that file and the number will be wrong for the rest of your code.

If you are supposed to keep these numbers together with the names and write out the calculations again, you'll have to do some work keeping the names together.

Here is an alternative version to solve the same task:

outputfile = raw_input("What is the output filename? :")

with open('Lab8.txt') as infile, open(outputfile, 'w') as outtfile:
    for name in infile:
        scores = next(infile).split()  # next() grabs the next line from infile here
        scores = map(int, scores)

        quiztotal = sum(scores[:4]) / 5
        midtermtotal = sum(scores[5:7]) / 2
        finaltotal = scores[7]
        score = quiztotal * .3 + midtermtotal * .4 + finaltotal * .3

        outfile.write(name)
        outfile.write('{0:0.2f}\n'.format(score))

pass is a statement, but inline if, being an operator, needs its operands to be expressions. map can’t actually remove elements from the sequence, but filter (returns a new list with only the values for which the function returns True) can:

print filter(lambda x: str(x)[-1] == '2', even)

I think you can use this:

pool = Pool(processes=5)
pool.map(main_func, [i for i in data.id.unique() if i in group.groups.keys()])

You don't need a lambda function for what you want to achieve. Try this:

>>> df["role"].str.split("/").str[0]
0          admin
1          admin
2    programming
Name: role, dtype: object

If for whatever reason you still want to use lambda, do this:

>>> df["role"].apply(lambda x: x.split("/")[0])
0          admin
1          admin
2    programming
Name: role, dtype: object