我正在尝试通过将元组的第一项与指定的int()进行比较来打印元组的列表,但是由于某些神秘的原因,第二项中值为0的元组才消失了
代码:
n_vertical = 3
n_horizontal = 3
for x in range(0,n_vertical):
for y in range(0,n_horizontal):
area.append((x,y,0))
print(area)
print('''
''')
def representacao_do_mapa(modo):
if modo == 1:
n=0
l_c = []
for x in area:
if x[0] == n:
l_c.append(x)
else:
print(l_c)
l_c = []
n+=1
representacao_do_mapa(1)
文本输出:
[(0, 0, 0), (0, 1, 0), (0, 2, 0)]
[(1, 1, 0), (1, 2, 0)]
答案 0 :(得分:3)
问题
创建新列表时,您将丢弃元组x
:
if x[0] == n: l_c.append(x) # here you append x else: print(l_c) # here you print but do nothing with x l_c = [] # list empty, x is missing n+=1
解决方案
def representacao_do_mapa(modo):
if modo == 1:
n=0
l_c = []
for x in area:
if x[0] == n:
l_c.append(x)
else:
print(l_c)
l_c = [x] # fix here
n+=1
print(l_c) # fix here
representacao_do_mapa(1)
除此之外-您的最后一个列表将不会被打印,因为最后一个l_c
永远不会进入代码的打印部分-您必须在该for-loop区域添加该标记。
输出(用于n_vertical = 3
和n_horizontal = 3
:
[(0, 0, 0), (0, 1, 0), (0, 2, 0)]
[(1, 0, 0), (1, 1, 0), (1, 2, 0)]
[(2, 0, 0), (2, 1, 0), (2, 2, 0)]
优化:
您可以使用列表推导和列表分解来缩短代码:
n_vertical = 3
n_horizontal = 3
area = [ (x,y,0) for x in range(n_horizontal) for y in range(n_vertical )]
# create with inner lists
area2 = [ [(x,y,0) for x in range(n_horizontal)] for y in range(n_vertical)]
print(area)
# print each inner list on new line
print(*area2, sep="\n")
或者您可以直接从area
打印:
print(* (area[i*n_horizontal:i*n_horizontal+n_horizontal]
for i in range(n_vertical)) , sep="\n")
使用生成器表达式将area
切成n_horizontal
个片段。
有关生成器/列表表达式的更多信息:Generator Expressions vs. List Comprehension
有关分块列表的更多信息:How do you split a list into evenly sized chunks?
更多关于列表切片的信息:Understanding slice notation
有关打印的更多信息:https://docs.python.org/3/library/functions.html#print