for i in range(10,100):
list_2.append(i)
for i in range(100,1000):
list_3.append(i)
for i in list_2:
str1 = str(i)
for j in range(2):
if int(str1[j]) == 0:
list_2.remove(i)
for i in list_3:
str2 = str(i)
for j in range(3):
if int(str2[j]) == 0:
list_3.remove(i)
I'm going to make a two-digit list and a three-digit list, and I'm going to delete 10, 20, 100, 200, and so on The two-digit list is deleted normally, but why is it not in list error in list_3.remove(i) in the three-digit list?
python list
for i in list_3:
str2 = str(i)
for j in range(3):
if int(str2[j]) == 0:
list_3.remove(i)
break
If you look at the 2nd list_3 logic, Assuming that 100 (hereinafter, list_3[0]) has been entered
1 > Non-zero if door not clear
Delete list_3[0] because 0 > 0.
Delete list_3[0] because 0 > 0 (already..) Error occurred because list_3[0] is not present)
Therefore
1 > Non-zero if door not clear
Delete list_3[0] because 0 > 0. break
You have to do it.
To make some improvements,
for i in list_2:
str1 = str(i)
#for j in range(2):
# # if int(str1[j]) == 0:
# # list_2.remove(i)
if "0" in str1:
list_2.remove(i)
To improve further,
list_2 = [ e for e in list_2 if "0" not in str(e) ]
I think the party is over, but there's also a way to solve it mathematically.
import math
# Basic idea: 2070% 100 == 70 == 2070% 1000
# This case is true because the third digit is zero.
def has_zero(num) :
# 10, 100, 1000... Divide it by and see if the rest of the previous one and the rest of the current one are the same.
lastMOD = 0
unit = 10;
while num >= unit :
thisMOD = num % unit
if thisMOD == lastMOD :
return True
lastMOD = thisMOD
unit = unit * 10;
return False
# It works amazingly.
print(has_zero(10))
print(has_zero(207))
print(has_zero(6000))
print(has_zero(70060))
print(has_zero(77767))
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