python - manipulate list comprehension -

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i have 3 lists make condition, if b equal 1 want find first element in c below a, can 2 indexes subtract them , time.

    = [1.373, 1.374, 1.374, 1.385, 1.385, 1.385, 1.374, 1.374]     b = [0, 1, 1, 0, 0, 0, 0, 0]     c = [0, 1.384, 1.385, 1.377, 0, 0, 0, 0]     view = [(nx, x, nz, z, nl, l) nx, x in enumerate(a)            nl, l in enumerate(b) nz, z in enumerate(c)            if (l == 1) & (nz <= nx) & (z > 0) & (z <= x) & (nz == nl)]

i created simple view list shows parameters better understanding. when numbers when c lower , want first elements... expected see [(3, 1.385, 1, 1.384, 1, 1), (3, 1.385, 2, 1.385, 2, 1)] , got that:

[(3, 1.385, 1, 1.384, 1, 1), (3, 1.385, 2, 1.385, 2, 1), (4, 1.385, 1, 1.384, 1, 1), (4, 1.385, 2, 1.385, 2, 1), (5, 1.385, 1, 1.384, 1, 1), (5, 1.385, 2, 1.385, 2, 1)]  time = [(nx - nz) nx, x in enumerate(a)         nl, l in enumerate(b) nz, z in enumerate(c)         if (l == 1) & (nz <= nx) & (z > 0) & (z <= x) & (nz == nl)]

how can first elements when condition true?

a solve without list comprehension , @ better time complexity:

a = [1.373, 1.374, 1.374, 1.385, 1.385, 1.385, 1.374, 1.374] b = [0, 1, 1, 0, 0, 0, 0, 0] c = [0, 1.384, 1.385, 1.377, 0, 0, 0, 0]  y_i, y in enumerate(b):     if y == 1:         x_i, x in enumerate(a):             if x >= c[y_i]:                 # prints 0 based indexes                 print "iteration {}, x: {}, x index: {} c: {}, c index {}".format(y_i, x, x_i, c[y_i], y_i)                 break

output:

iteration 1, x: 1.385, x index: 3 c: 1.384, c index 1 iteration 2, x: 1.385, x index: 3 c: 1.385, c index 2

sometimes it;s easier start normal loops when algorithm can make better code, there no need 3 statements.


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