For a pair of spike trains, TA (train a) and TB, these related routines return the count of the number of spikes in B that occur within some time window [-tmax,tmax] of a spike in A. For histbi.ab, we return a histogram of dt values from [-tmax,tmax]. For hist.ab, we ignore the sign of each dt and just return a histogram in the range [0,tmax]. Finally, for count.nab, we just return the number of dt values found in the range [-tmax,tmax], rather than binning dt into a histogram.

1 2 3 |

`ta` |
Vector of spike times, sorted such that lowest is first. |

`tb` |
Vector of spike times, sorted such that lowest is first. |

`tmax` |
maximum time (in seconds) to bin |

`nbins` |
Number of bins in the histogram. For histbi.ab, each bin is of width (2*tmax)/nbins. For hist.ab, each bin is (tmax)/nbins wide. |

`hist.ab`

returns a histogram of counts ignoring sign.
`histbi.ab`

returns a histogram of counts respecting sign.
`count.nab`

returns the number of dt values.

For the histogram routines, each bin is of the form [low, high), with the exception of the last bin (for +tmax), which is of the form [tmax-binwid, tmax]. By assuming the spikes are ordered lowest first, the number of spike comparisons is greatly reduced, rather than comparing each spike with A with each spike in B.

No references here.

Nothing else yet...

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 | ```
stopifnot(isTRUE(all.equal.numeric(
histbi.ab(c(0), c(-2, -2, 0, 0, 1, 1,1, 1.8,2), tmax=2, nbins=4),
c(2,0,2,5),
check.attributes=FALSE)))
stopifnot(identical(TRUE, all.equal.numeric(
hist.ab(c(0), c(-2, -2, 0, 0, 1, 1,1, 1.8, 2), tmax=2, nbins=4),
c(2,0,3,4),
check.attributes=FALSE)))
test.hist.ab()
## Not run:
test.histograms.versus.r()
test.count.hist.nab()
test.count.hist.nab(s)
test.count.hist2.nab(s)
## End(Not run)
``` |

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