ask how the variables of the expression change, and how the results, especially n, change.
fun = [(lambda n: i + n) for i in range (10)]
ask how the variables of the expression change, and how the results, especially n, change.
fun = [(lambda n: i + n) for i in range (10)]
this is one of the few "strange things" in python
fun = [(lambda n: i+n) for i in range(10)]
-sharp
fun = [(lambda n: 9+n) for _ in range(10)]because the parameter I in lambda is determined at run time, not at declaration time.
finally fun saves an array of 10 elements, each of which is a lambda method, which is equal to the following lambda_item function
def lambda_item(n):
return 9 + nif the entire expression is interpreted in python code, it can be like this
fun = []
for i in range(10):
fun.append((lambda n: 9+n))another thing that confuses newcomers is
a = [[0]] * 3
a[0].append(1)
print(a)print out
[[0,1], [0,1], [0,1]]first of all, I would like to thank @ Li Yi for his answer. His answer is very good.
so, what is the result of this program?
-sharp fun = [(lambda n : i + n) for i in range(10)]
funs = []
tmp = 0
fun = lambda n: tmp + n
for i in range(10):
funs.append(fun)
tmp = i
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