List comprehension with complex conditions in python

Question

I was looking up for ways to make my loop fast,then I found about list comprehensions.

I tried it on my own, but I don't fully understand it yet.

From what I learned researching about list comprehensions, the code I like to execute would be on the left side, followed by the conditions then the for loop.

So, it would basically look like this.

["Something I'd like to execute" Some conditions for loop]
Following this style, I did it like this.

The code I was trying to turn into a one liner:

graph = []

for g in range(M):
    satisfy = []
    graph_count = 0
    for i in range(N-1):
        count = 0
        for j in range(N):
            if i < j and count < 1:
                if graph_count < g:
                    count += 1
                    graph_count += 1
                    satisfy.append("1")
                else:
                    satisfy.append("0")
            elif i < j:
                satisfy.append("0")
    graph.append("".join(map(str,satisfy)))

My Attempt

graph = [[count+=1,graph_count+=1,satisfy.append("1") if graph_count < g else satisfy.append("0") and if i<j and count<1 else satisfy.append("0") if i<j for j in range(N) count=0 for i in range(N-1)] graph_count=0, "".join(map(str,satisfy)) for g in range(M)]

What am I doing wrong?

Answer 1

There are two kinds of optimization:

• micro-optimization (statement level, e.g. f-strings are faster than format function)
• macro optimization (algorithm, used data structures, etc)

Optimizing algorithms may have much higher returns than spending the same effort on micro-optimizations.

Here is my solution to your problem. After analyzing the expected output, I found some patterns and reduced the number of loops and allocations:

from itertools import accumulate

graph = []
size = N * (N - 1) // 2
s = list(reversed(list(accumulate(range(N)))))
for g in range(M):
    satisfy = ["0"] * size
    for i in range(N - 1):
        if g > i:
            satisfy[size - s[i]] = "1"

    graph.append("".join(satisfy))