没有索引的python循环

Question

我正在学习 python，我喜欢你可以循环遍历可迭代对象而无需不断创建索引变量。

但是我发现自己一直在创建索引变量，所以我可以在来自并行对象的调用中引用它们。就像我比较两个列表或一个列表和一个字典时一样。

根据要求：一些例子

`s= "GAGCCTACTAACGGGAT"# the strings we are evaluating
t= "CATCGTAATGACGGCCT"

c = 0# c = position of interest within the strings
m = 0 # m = number of mutations found so far. 

for i in s: # cycling through all nucleotides listed in s
if(i == t[c]): #compare s[c] to t[c]
    c +=1 # if equal, increase position counter
else:
    c+=1 # if not equal, increase both position and                 
    m+=1        #mutation counters.`

与

def allPossibleSubStr(s): #takes a dict of strings, finds the shortest, and produces a list of all possible substrings, thus a list of possible common substrings to all elements of the original dict
ks = s.keys()
c=0 #counter
j=0 # ks place holder for shortest string
subSTR = []

for i in ks: #finds the shortest entry in stringDict
    if(s[i] < s[ks[(c+1)%len(s)]]):
        j=c
    c +=1

c=s[ks[j]] #c is now a string, the shortest... 
j=ks[j] # j is now a key string, the shortest...
n = (len(c)*(len(c)+1))/2 # number of subsets of c

#producing a list of possible substrings
for i in range(len(c)):
    for k in range(len(c)-i):
        subSTR.append(c[i:i+k+1])
        #print("i =" +str(i)+ " and k=" + str(k))
#is there a list function with eleminates duplicate entries. 
subSTR=list(set(subSTR))# a set does not have any duplicate entires
subSTR.sort(key=len) # sorts substring from shortest to longest string
subSTR.reverse()

return subSTR

有没有办法解决这个问题？

Answer 1

听起来你写的代码是这样的：

for i, item in enumerate(list1):
    if item == list2[i]:
        ...

您仍然不需要显式索引，因为您可以压缩两个列表并遍历元组列表。

for item1, item2 in zip(list1, list2):
    if item1 == item2:

你可以用两个字典做同样的事情，虽然因为它们是无序的，你可能想在排序后压缩它们的键：

for key1, key2 in zip(sorted(dict1), sorted(dict2)):

Answer 2

使用枚举...

for i, item in enumerate(myList):
  foo = parallelList[i]

或压缩...

for item, foo in zip(myList, parallelList):
  ...