Computation of Only Even or Odd Frequency Bins of DFT
Given a signal $x[n]$ of length N, assume $X[k]$ is its N-point DFT $$X[k] = \sum_{n=0}^{N-1}{x[n]e^{-j\frac{2\pi}{N}nk}}$$ where $k = 0, 1,..., N-1$  computed via N-point FFT. And if we wish to compute separateley those even and odd indexed samples of the DFT $X[k]$ , we can proceed with the following:

Consider those even indexed samples of $X[k]$ which are denoted as $X_e[k]$ where:
\begin{align}
X_e[k] = X[2k] &amp;=  \sum_{n=0}^{N-1}{x[n]e^{-j\frac{2\pi}{N}n(2k)}} &amp;\scriptstyle{\text{term 2k computes even samples of X[k]}}\\
X_e[k] &amp;=  \sum_{n=0}^{N-1}{x[n]e^{-j\frac{2\pi}{N/2}nk}} &amp;\scriptstyle{\text{factor 2 is moved into N/2 term}}\\
X_e[k] &amp;=  \sum_{n=0}^{\frac{N}{2}-1}{x[n]e^{-j\frac{2\pi}{N/2}nk}} + \sum_{n=\frac{N}{2}}^{N-1}{x[n]e^{-j\frac{2\pi}{N/2}nk}} &amp;\scriptstyle{\text{Divide the sum into 2 halves}}\\
X_e[k] &amp;=  \sum_{n=0}^{\frac{N}{2}-1}{x[n]e^{-j\frac{2\pi}{N/2}nk}} + \sum_{n=0}^{\frac{N}{2}-1}{x[n+\frac{N}{2}]e^{-j\frac{2\pi}{N/2}nk}} &amp;\scriptstyle{\text{adjust the 2nd sum range}}\\
X_e[k] &amp;=  \sum_{n=0}^{\frac{N}{2}-1}{ (x[n]+x[n+\frac{N}{2}])e^{-j\frac{2\pi}{N/2}nk}}  &amp;\scriptstyle{\text{merge the 2 sums}}\\
\end{align}

The final form suggest that the required even samples of N-point DFT of signal $x[n]$ can be computed from an N/2 point DFT of a new signal $x_{half}[n]= x[n]+x[n+ N/2]$ whose length is half that of $x[n]$  

The signal $x_{half}[n]$ is simply computed by adding the first half of the signal $x[n]$ into its second half. (if the length N of signal x[n] is not even padd a zero to its tail to make it even) 

In short the following matlab code gives you the desired even samples of X[k] without computing a full N point DFT (via FFT) of x[n]: 

Xe = fft ( x(1:N/2) + x(N/2 + 1 : end), N/2);


Note that because of the addition of those half length signals efficiency will degrade from a simple N/2 point FFT.

The case for the odd indexed samples proceeds similary but results in a more complex form which I would like to summarize in the following Matlab script:

% Let x[n] be the signal of length N
xc = x .* exp(-j*2*pi*[0:N-1]/N) ; % Multiply x[n] with a complex phase term
Xo = fft (xc(1:N/2) + xc(N/2 + 1, N), N/2); % odd indexed samples of X[k]


Due to a complex multiplication before the FFT, in this case of computing odd indices performance is further reduced but still preferable over a direct N point FFT.

The following Matlab excerpt demonstrates both even and odd samples (with odds requiring only half the complex multiplication wrt the above line)

N = 1024;
x = randn(1,N);
X = fft(x,N);       % original DFT of X

Xe = fft( x(1:N/2) + x(N/2 + 1: N) , N/2);                          % Even   samples only from N/2 point FFT
Xo = fft( ( x(1:N/2) - x(N/2+1:N) ).*exp(-j*pi*[0:2:N-2]/N) , N/2); % Odd samples only from N/2 point FFT

figure,stem( abs ( Xe - X(1:2:N)) ); title('Xe - X(1:2:N)');
figure,stem( abs ( Xo - X(2:2:N)) ); title('Xo - X(2:2:N)');


Note the absolute value of the difference. It is neither zero (down to roundoff error), nor too big. But for the purpose of particular computation this "practical" error shall be considered carefully.