在python中，['a'] * 4给出['a', 'a', 'a', 'a']。为什么不[['a'], ['a'], ['a'], ['a']]？
答案 0 :(得分：5)
因为，就像重复整数乘法一样，用lis乘法也是重复串联。
>>> ['a'] + ['a'] + ['a'] + ['a']
['a', 'a', 'a', 'a']
>>> ['a'] * 4
['a', 'a', 'a', 'a']
>>> 4 * ['a']
['a', 'a', 'a', 'a']
这也遵守其他各种法律：
lst * n + lst * m == lst * (n + m) lst * 0 == 0 * lst == [] lst * 1 == lst。以上内容也与空白列表[]为加成身份一致，
['a'] + [] == [] + ['a'] == ['a']
因为
['a'] + [] == ['a'] * 1 + ['a'] * 0 # zero property and multiplicative identity
== ['a'] * (1 + 0) # distribution
== ['a'] * 1 # integer addition
== ['a'] # multiplicative identity
可以说，list * int不应在 all 处定义。为了快速生成n个相同元素的列表，主要用例是使用列表推导（稍后添加到语言中）更安全地完成。 [x] * n创建对单个元素x的引用列表，如果x可变，通常不是。另一方面，[x for _ in range(n)]创建由x定义的独立对象的列表，当x可变时，就是您想要的。
答案 1 :(得分：1)
类似于与数字carpet_remaining = 2
if (carpet_remaining > 0):
for i in range(100):
newQ,newT = currentValues()
print(newQ)
print(newT)
# Attempt to start a mission:
max_retries = 3
for retry in range(max_retries):
try:
agent_host.startMission( my_mission, my_mission_record )
break
except RuntimeError as e:
if retry == max_retries - 1:
print("Error starting mission:",e)
exit(1)
else:
time.sleep(2)
# Loop until mission starts:
print("Waiting for the mission to start ", end=' ')
world_state = agent_host.getWorldState()
while not world_state.has_mission_begun:
print(".", end="")
time.sleep(0.1)
world_state = agent_host.getWorldState()
for error in world_state.errors:
print("Error:",error.text)
print()
print("Mission running ", end=' ')
agent_host.sendCommand("pitch 0") #Start looking downward slowly
time.sleep(1) #Wait a second until we are looking in roughly the right direction
while world_state.is_mission_running: #may have to restart mission. look into it
print(".", end="")
time.sleep(0.1)
world_state = agent_host.getWorldState()
for error in world_state.errors:
print("Error:",error.text)
if world_state.number_of_observations_since_last_state > 0:
msg = world_state.observations[-1].text
observations = json.loads(msg).get("LineOfSight")
#print(observations)
# 1. Determine block type
#2. Make a switch case based on block type that determines to set input nodes.
# for example if block type is 'grass' then grass node is 1, and all other nodes 0.
# 3. Run forward propagation.
# 4. Look at output nodes figure out node with highest value
# 5. execute action based on output nodes
#botsEyes = observations.get("LineOfSight")
#print (botsEyes)
botsEyes = observations.get("type")
# print('The bot is looking at: ',botsEyes)
if botsEyes == ("lapis_block"):
inputValues = [1, 0, 0, 0]
lapisNode = 1
grassNode = 0
fenceNode = 0
carpetNode = 0
elif botsEyes == ("grass"):
inputValues = [0, 1, 0, 0]
lapisNode = 0
grassNode = 1
fenceNode = 0
carpetNode = 0
elif botsEyes == ("spruce_fence"):
inputValues = [0, 0, 1, 0]
lapisNode = 0
grassNode = 0
fenceNode = 1
carpetNode = 0
elif botsEyes == ("carpet"):
inputValues = [0, 0, 0, 1]
lapisNode = 0
grassNode = 0
fenceNode = 0
carpetNode = 1
else:
inputValues = [0, 0, 0, 0]
lapisNode = 0
grassNode = 0
fenceNode = 0
carpetNode = 0
print("Neural Net values: ",inputValues)
X = np.array((inputValues,), dtype = float)
#print ("Actual Output: \n" + str(y))
#print ("Predicted Output: \n" + str(NN.forward(X)))
#print ("Loss: \n" + str(np.mean(np.square(y - NN.forward(X))))) # mean sum squared loss
o = NN.forward(X)
# print ("Actual Output: \n" + str(o))
largestValue = np.amax(o)
print("Highest Value:" , largestValue)
myaction,myreward = getAction()
if myaction == "use 1": #place
carpet_remaining = carpet_remaining - 1
print("*********** losing carpet ****************")
print(carpet_remaining)
的乘法等效于['a'] * 4，因此也就是['a'] + ['a'] + ['a'] + ['a']。
答案 2 :(得分：0)
['a'] * 4 == ['a'] + ['a'] + ['a'] + ['a']
['a'] + ['a'] = [['a'],['a']].
['a'] + ['a'] + ['a'] = [[['a'],['a']],['a']] #??
['a'] + ['a'] + ['a'] + ['a'] = [[[['a'],['a']],['a']],['a']] # ?????????
(['a']*2)*3 # what should be output of this?