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私は2つのリストを持っています。 list1の各点から始めて、list2に最も近い後続の(大きい)時間を見つけたいと思います。Pythonの交換時間を見つける最速の方法
LIST1 = [280、290]
LIST2 = [282、295]
交換(LIST1、LIST2)= [2,5]
I:例えば
これをすばやくやるのに困っている。私が考えることができると思う唯一の方法は、list1の各要素をループし、list1要素よりもリストyの最初のヒットを取ることです(リストがソートされます)。私の下の2つの試み一匹のパンダ一W/Oパンダ:
# dictionary containing my two lists
transition_trj = {'ALA19': [270.0, 280.0, 320.0, 330.0, 440.0, 450.0,
470.0], 'ALA88': [275.0, 285.0, 325.0, 333.0, 445.0, 455.0, 478.0]}
# for example, exchange times for ('ALA19','ALA88') = [5.0, 5.0, 5.0, 3.0, 5.0, 5.0, 8.0]
#find all possible combinations
names = list(transition_trj.keys())
import itertools
name_pairs = list(itertools.combinations_with_replacement(names, 2))
# non-pandas loop, takes 1.59 s
def exchange(Xk,Yk): # for example, a = 'phiALA18', b = 'phiARG11'
Xv = transition_trj[Xk]
Yv = transition_trj[Yk]
pair = tuple([Xk,Yk])
XY_exchange = [] # one for each pair
for x in range(len(Yv)-1): # over all transitions in Y
ypoint = Yv[x] # y point
greater_xpoints = []
for mini in Xv:
if mini > ypoint:
greater_xpoints.append(mini) # first hit=minimum in sorted list
break
if len(greater_xpoints) > 0:
exchange = greater_xpoints[0] - ypoint
XY_exchange.append(exchange)
ET = sum(XY_exchange) * (1/observation_t)
return pair, ET
# pandas loop, does same thing, takes 11.58 s...I am new to pandas...
import pandas as pd
df = pd.DataFrame(data=transition_trj)
def exchange(dihx, dihy):
pair = tuple([dihx, dihy])
exchange_times = []
for x in range(df.__len__()):
xpoint = df.loc[x, dihx]
for y in range(df.__len__()):
ypoint = df.loc[y, dihy]
if ypoint > xpoint:
exchange = ypoint - xpoint
exchange_times.append(exchange)
break
ET = sum(exchange_times) * (1/observation_t)
return pair, ET
# here's where I call the def, just for context.
exchange_times = {}
for nm in name_pairs:
pair, ET = exchange(nm[0],nm[1])
exchange_times[pair] = ET
if nm[0] != nm[1]:
pair2, ET2 = exchange(nm[1], nm[0])
exchange_times[pair2] = ET2
、試してください:[ 'ALA88'] REINDEX(POS).reset_index(ドロップ=真)DF - [ 'ALA19']あなたがいる – Leigh
DF。それはよりクリーンで速いです。 –