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Python 2.7 min函数只打印一个结果,而不是两个结果_Python 2.7_Min - Fatal编程技术网

Python 2.7 min函数只打印一个结果,而不是两个结果

Python 2.7 min函数只打印一个结果,而不是两个结果,python-2.7,min,Python 2.7,Min,“”“这是我的代码。(向第一次使用此代码的用户道歉),但它不能满足我的要求。如果运行此代码,最小值为-2,并且有两个键带有-2,因此位置应该有两个结果,而不是一个。有人能解释为什么会发生这种情况吗?”第一个索引以0开头 def Skew(Text): skew = {} n = len(Text) skew[0] = 0 for i in range(1,n+1): #Every time we encounter a G, skew

“”“这是我的代码。(向第一次使用此代码的用户道歉),但它不能满足我的要求。如果运行此代码,最小值为-2,并且有两个键带有-2,因此位置应该有两个结果,而不是一个。有人能解释为什么会发生这种情况吗?”

第一个索引以0开头

def Skew(Text):    
    skew = {}
    n = len(Text)
    skew[0] = 0
    for i in range(1,n+1):   
        #Every time we encounter a G, skew[i] is equal to skew[i-1]+1
        if Text[i-1] == "G": skew[i] = skew[i-1]+1
        #every time we encounter a C, Skew[i] is equal to Skew[i-1]-1
        elif Text[i-1] == "C": skew[i] = skew[i-1]-1
        #otherwise, Skew[i] is equal to Skew[i-1]
        else: skew[i] = skew[i-1]
return skew


Text = "CCGGCCGG"    
positions = [] #output variable
skew = Skew(Text)
print skew
minimum = min(skew.values()) 
print minimum
#use the for loop, to look for i when Skew[i]=minimum
for i in skew:
    if skew[i] == minimum: positions.append(i)
positions = positions[1:]
print positions

这是因为您删除了position中的第一个结果,该行为:positions=positions[1:]。@PiotrDabkowski。我现在明白了。谢谢


    positions = positions[0:]