Again, many thanks to all who provide their solution.
I have timed these (though on my old P3(0.9GHz)) - see below
Helmut.
Helmut Jarausch wrote:
Hi,
I'm looking for an elegant solution of the following tiny but common
problem.
I have a list of tuples (Unique_ID,Date) both of which are strings.
I want to delete the tuple (element) with a given Unique_ID, but
I don't known the corresponding Date.
#!/usr/bin/python
import random
import timeit
Lorg=[]
def genList(L) :
for f in range(ord('A'),ord('z')+1) :
for s in range(ord('A'),ord('z')+1) :
L.append((chr(f)+chr(s),str(random.randrange(0,100
genList(Lorg)
Times= 1000
T0= timeit.Timer('L=list(Lorg)','from __main__ import Lorg').timeit(Times)
print T0
SetUp1=r'''from __main__ import Lorg
def del_by_key(L,key) :
d= dict(L)
del d[key]
L[:]= d.items()
'''
SetUp2=r'''from __main__ import Lorg
def del_by_key(L,key) :
d= dict(L)
t= (key,d[key])
L.remove(t)
'''
SetUp3=r'''from __main__ import Lorg
def del_by_key(L,key) :
index= [k for k,val in L]
pos = index.index(key)
del L[pos]
'''
SetUp4=r'''from __main__ import Lorg
def del_by_key(L,key) :
for pos, (k,d) in enumerate(L):
if k == key :
del L[pos]
break
'''
SetUp5=r'''from __main__ import Lorg
def del_by_key(L,key) :
L[:]= [(k,d) for (k,d) in L if k !=key]
'''
SetUp6=r'''from __main__ import Lorg
class Tester(object) :
def __init__(self,key) :
self.key= key
def __eq__(self,other) :
return other[0] == self.key
def del_by_key(L,key) :
del L[L.index(Tester(key))]
'''
print '*** ready ***'
T= timeit.Timer(L=list(Lorg);del_by_key(L,'Zz'),SetUp1).timeit(Times)
print Method 1 :,T-T0
T= timeit.Timer(L=list(Lorg);del_by_key(L,'Zz'),SetUp2).timeit(Times)
print Method 2 :,T-T0
T= timeit.Timer(L=list(Lorg);del_by_key(L,'Zz'),SetUp3).timeit(Times)
print Method 3 :,T-T0
T= timeit.Timer(L=list(Lorg);del_by_key(L,'Zz'),SetUp4).timeit(Times)
print Method 4 :,T-T0
T= timeit.Timer(L=list(Lorg);del_by_key(L,'Zz'),SetUp5).timeit(Times)
print Method 5 :,T-T0
T= timeit.Timer(L=list(Lorg);del_by_key(L,'Zz'),SetUp6).timeit(Times)
print Method 6 :,T-T0
# Results on an old P3 (0.9 GHz)
# *** ready ***
# Method 1 : 10.9850928783
# Method 2 : 5.96455168724
# Method 3 : 3.97821164131
# Method 4 : 1.66151881218
# Method 5 : 8.90886187553
# Method 6 : 6.2503888607
The clear winner is
def del_by_key(L,key) :
for pos, (k,d) in enumerate(L):
if k == key :
del L[pos]
break
--
Helmut Jarausch
Lehrstuhl fuer Numerische Mathematik
RWTH - Aachen University
D 52056 Aachen, Germany
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