[英]Replacing a vertical sublist in a list of lists
該問題是對該問題的擴展。
我使用列表L
表示二維數組,例如:
[ [1, 2, 3, 4],
[1, 2, 3, 4],
[1, 2, 3, 4],
[1, 2, 3, 4] ]
對於給定的子列表,例如[9, 99]
,我想使用類似以下的直觀方式用該sublist
替換“ 2-D”列表中的特定子列表:
L[1][0:2] = sublist
# which updates `L` to:
[ [1, 2, 3, 4],
[1, 9, 99, 4],
[1, 2, 3, 4],
[1, 2, 3, 4] ] # not in this format, but written like this for clarity
這適用於水平替換,但不適用於垂直替換,因為我們不能像這樣分割列表: L[0:2][0]
。 如果必須使用此切片系統,則可以轉置L
( 列表的 轉置 列表 ),然后使用此切片方法,然后將其轉回。 但是,即使是為了簡單起見,這也不是很有效。
復制 L[0:2][0]
並獲得此輸出 的有效方法是 什么?
[ [1, 2, 3, 4],
[1, 9, 3, 4],
[1, 99, 3, 4],
[1, 2, 3, 4] ]
注意:假設len(sublist) <= len(L)
,以進行垂直替換(這是此問題的重點)。
循環方法:
def replaceVert(al : list, repl:list, oIdx:int, iIdx:int):
for pos in range(len(repl)):
al[oIdx+pos][iIdx] = repl[pos]
a = [ [1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12],
[13, 14, 15, 16] ]
print(a) # [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]
replaceVert(a,['ä','ü'],2,2) # this is a one liner ;)
print(a) # [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 'ä', 12], [13, 14, 'ü', 16]]
轉置/切片/轉置方法:
我完全忽略了“不移調”的提法。 這是使用Q不需要的帶有切片的轉置,更改,轉置方法。 這是該問題標題的答案,因此我決定將其留給以后的人們進行搜索,然后迷失於此問題:
a = [ [1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12],
[13, 14, 15, 16] ]
b = list(map(list,zip(*a))) # will make [ [1,5,9,13], ... ,[4,8,12,16]]
b[1][0:2]=['a','b'] # replaces what you want here (using a and b for clarity)
c = list(map(list,zip(*b))) # inverts b back to a's form
print(a)
print(b)
print(c)
輸出:
[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]] # a
[[1, 5, 9, 13], ['a', 'b', 10, 14], [3, 7, 11, 15], [4, 8, 12, 16]] # b replaced
[[1, 'a', 3, 4], [5, 'b', 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]] # c
定時4x4列表,2個替換 :
setuptxt = """
def replaceVert(al : list, repl:list, oIdx:int, iIdx:int):
for pos in range(len(repl)):
al[oIdx+pos][iIdx] = repl[pos]
a = [ [1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12],
[13, 14, 15, 16] ]
"""
zipp = """b = list(map(list,zip(*a)))
b[1][0:2]=['a','b']
c = list(map(list,zip(*b)))
"""
import timeit
print(timeit.timeit("replaceVert(a,['ä','ü'],2,2)",setup = setuptxt))
print(timeit.timeit(stmt=zipp, setup=setuptxt))
輸出:
looping: 12.450226907037592
zipping: 7.50479947070815
ZIPPing(轉置/切片/轉置)方法需要大約60%的時間用於4x4列表。
更大的列表替換了1000x1000和〜70個元素:
setuptxt = """
def replaceVert(al : list, repl:list, oIdx:int, iIdx:int):
for pos in range(len(repl)):
al[oIdx+pos][iIdx] = repl[pos]
a = [ [kk for kk in range(1+pp,1000+pp)] for pp in range(1,1000)]
repl = [chr(mm) for mm in range(32,100)]
"""
import timeit
print(timeit.timeit("replaceVert(a,repl,20,5)",number=500, setup = setuptxt))
zipp = """b = list(map(list,zip(*a)))
b[20][5:5+len(repl)]=repl
c = list(map(list,zip(*b)))
"""
print(timeit.timeit(stmt=zipp, setup=setuptxt,number=500))
輸出:
looping: 0.07702917579216137
zipping: 69.4807168493871
循環獲勝。 感謝@Sphinx的評論
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