Python, 13

If the OP wants all functions to be non-anonymous, add f= to the front of each for two additional characters.

lambda x:x//1

Since x%1 returns the amount following the decimal point, this is pretty short (14):

lambda x:x-x%1

If using cast to integer instead, it's one character longer (15):

lambda x:int(x)

The shortest using string casting I could come up with (40):

lambda x:int(`x`.split('.')[0])+cmp(x,0)

Python, 13

If the OP wants all functions to be non-anonymous, add f= to the front of each for two additional characters.

lambda x:x//1

Since x%1 returns the amount following the decimal point, this is pretty short (14):

lambda x:x-x%1

The shortest using string casting I could come up with (40):

lambda x:int(`x`.split('.')[0])+cmp(x,0)

#Python, 13

If the OP wants all functions to be non-anonymous, add f= to the front of each for two additional characters.

lambda x:x//1

Since x%1 returns the amount following the decimal point, this is pretty short (14):

lambda x:x-x%1

If using cast to integer instead, it's one character longer (15):

lambda x:int(x)

The shortest using string casting I could come up with (40):

lambda x:int(`x`.split('.')[0])+cmp(x,0)

Python, 13

If the OP wants all functions to be non-anonymous, add f= to the front of each for two additional characters.

lambda x:x//1

Since x%1 returns the amount following the decimal point, this is pretty short (14):

lambda x:x-x%1

If using cast to integer instead, it's one character longer (15):

lambda x:int(x)

The shortest using string casting I could come up with (40):

lambda x:int(`x`.split('.')[0])+cmp(x,0)

#Python, 13

If the OP wants all functions to be non-anonymous, add f= to the front of each for two additional characters.

lambda x:x//1

Since x%1 returns the amount following the decimal point, this is pretty short (14).

lambda x:x-x%1

If using cast to integer instead, it's one character longer than using modulus (15).

lambda x:int(x)

#Python, 13

If the OP wants all functions to be non-anonymous, add f= to the front of each for two additional characters.

lambda x:x//1

Since x%1 returns the amount following the decimal point, this is pretty short (14):

lambda x:x-x%1

If using cast to integer instead, it's one character longer (15):

lambda x:int(x)

The shortest using string casting I could come up with (40):

lambda x:int(`x`.split('.')[0])+cmp(x,0)