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Python 用于计算元素的嵌套while循环有什么问题

python python-3.x

Python 用于计算元素的嵌套while循环有什么问题,python,python-3.x,Python,Python 3.x,我只想计算出F的每个元素在N中出现的频率，然后打印出来。我使用了嵌套for循环，它可以工作。但当我使用嵌套的while循环时，它并没有像预期的那样工作。我检查了代码，但找不到原因 F = [4,7,2] N = [2,3,4,2,5,6,3,2,6,7,3,4] 嵌套循环版本，按预期工作： four_count = 0 seven_count = 0 two_count = 0 for n in N: for f in F: if n == f and f == 4:

我只想计算出F的每个元素在N中出现的频率，然后打印出来。我使用了嵌套for循环，它可以工作。但当我使用嵌套的while循环时，它并没有像预期的那样工作。我检查了代码，但找不到原因

F = [4,7,2]
N = [2,3,4,2,5,6,3,2,6,7,3,4]
嵌套循环版本，按预期工作：

four_count = 0
seven_count = 0
two_count = 0

for n in N:
    for f in F:
        if n == f and f == 4:
            four_count += 1
        elif n == f and f == 7:
            seven_count += 1
        elif n == f and f == 2:
            two_count += 1

print(str(F[0]) + " occurs in N " + str(four_count) + " times")
print(str(F[1]) + " occurs in N " + str(seven_count) + " times")
print(str(F[2]) + " occurs in N " + str(two_count) + " times")

4 occurs in N 2 times
7 occurs in N 1 times
2 occurs in N 3 times
这是正确的输出：

four_count = 0
seven_count = 0
two_count = 0

for n in N:
    for f in F:
        if n == f and f == 4:
            four_count += 1
        elif n == f and f == 7:
            seven_count += 1
        elif n == f and f == 2:
            two_count += 1

print(str(F[0]) + " occurs in N " + str(four_count) + " times")
print(str(F[1]) + " occurs in N " + str(seven_count) + " times")
print(str(F[2]) + " occurs in N " + str(two_count) + " times")

4 occurs in N 2 times
7 occurs in N 1 times
2 occurs in N 3 times
嵌套while循环版本，输出错误：

four_count = 0
seven_count = 0
two_count = 0

N_Count = 0
F_Count = 0

while N_Count < len(N):
    while F_Count < len(F):
        if N[N_Count] == F[F_Count] and F[F_Count] == 4:
            four_count += 1
        elif N[N_Count] == F[F_Count] and F[F_Count] == 7:
            seven_count += 1
        elif N[N_Count] == F[F_Count] and F[F_Count] == 2:
            two_count += 1
        F_Count += 1
    N_Count += 1

print(str(F[0]) + " occurs in N " + str(four_count) + " times")
print(str(F[1]) + " occurs in N " + str(seven_count) + " times")
print(str(F[2]) + " occurs in N " + str(two_count) + " times")
当N\u Count时，必须在
之后重置
F\u Count=0
，否则list
F
只循环一次。因此，这将是：


...
while N_Count < len(N):
    F_Count = 0
    while F_Count < len(F):
...



或者类似的
当N\u Count
时，您必须在
之后重置
F\u Count=0
，否则list
F
只循环一次。因此，这将是：


...
while N_Count < len(N):
    F_Count = 0
    while F_Count < len(F):
...



或者类似的
你是对的。谢谢你救了我一天。“在while N_Count