![](/img/trans.png)
[英]Python: else statement returns True, though it's mentioned to return False
[英]Python : return True if … else False statement (Boolean Value)
A <-> G和C <-> T之間的過渡
A <-> C和G <-> T之間的轉換分配:
- 編寫一個需要兩個核苷酸的函數轉換。 該函數必須返回一個布爾值,該布爾值指示是否用第二個核苷酸替換第一個核苷酸導致轉換。
問題:我認為函數無法識別or
后面的語句。 該代碼在某些情況下不起作用,例如: transition('A', 'G')
為True,而對於我的代碼,他給出False
- 寫一個取兩個DNA序列s1和s2的函數比
該函數可以假定兩個序列的長度相同(該函數不需要顯式檢查此長度)。 該函數必須返回兩個給定序列的轉換/轉換比R(s 1,s 2)∈R作為浮點數。 在兩個序列之間不存在顛換的情況下,R(s 1,s 2)= 0。
問題:代碼無法正常工作
def transition(letter1, letter2):
"""
>>> transition('G', 'A')
True
>>> transition('t', 'g')
False
>>> transition('C', 'c')
False
"""
return True if letter1.lower() == 'gt' and letter2.lower() == 'ac' or letter1.lower() == 'ac' and letter2.lower() == 'gt' else False
def transversion(letter1, letter2):
"""
>>> transversion('G', 'A')
False
>>> transversion('t', 'g')
True
>>> transversion('C', 'c')
False
"""
return True if letter1.lower() == 'ct' and letter2.lower() == 'ag' or letter1.lower() == 'ag' and letter2.lower() == 'ct' else False
def ratio(seq1, seq2):
"""
>>> ratio('ATTAGCATTATCATC', 'AAATAGGATATATGG')
0.2222222222222222
>>> seq1 = 'GCAACGCACAACGAAAACCCTTAGGGACTGGATTATTTCGTGATCGTTGTAGTTATTGGAAGTACGGGCATCAACCCAGTT'
>>> seq2 = 'ttatctgacaaagaaagccgtcaacggctggataatttcgcgatcgtgctggttactggcggtacgagtgttcctttgggt'
>>> ratio(seq1, seq2)
1.2142857142857142
"""
count = 0
tel = 0
for i in range(len(seq1)):
if transition(seq1[i], seq2[i]):
count += 1
for i in range(len(seq1)):
if transversion(seq1[i], seq2[i]):
tel += 1
if tel != 0:
return float(count / tel)
else:
return 0
if __name__ == '__main__':
import doctest
doctest.testmod()
像這樣改變線條
return True if letter1.lower() == 'gt' and letter2.lower() == 'ac' or letter1.lower() == 'ac' and letter2.lower() == 'gt' else False
創建人:
return (letter1.lower() == 'gt' and letter2.lower() == 'ac' or letter1.lower() == 'ac' and letter2.lower() == 'gt')
我認為同一核鹼基的transition
或transversion
必須返回False
(即transition('A','A')
== False)
您可以使用簡單的命名謂詞來簡化操作: https : //repl.it/N4TC/4
def transition(nucleobase1, nucleobase2):
""" True if both are different and are purine
"""
return (not isEqual(nucleobase1, nucleobase2) and
isPurine(nucleobase1) and
isPurine(nucleobase2))
def transversion(nucleobase1, nucleobase2):
""" True if both are different and not transition
"""
return (not isEqual(nucleobase1, nucleobase2) and
not transition(nucleobase1, nucleobase2))
其他謂詞:
### nucleobase Predicat
def isAdenine(nucleobase):
""" True if adenine (A)
"""
return nucleobase.lower()=='a'
def isCytosine(nucleobase):
""" True if cytosine (C)
"""
return nucleobase.lower()=='c'
def isGuanine(nucleobase):
""" True if guanine (G)
"""
return nucleobase.lower()=='g'
def isThymine(nucleobase):
""" True if thymine (T)
"""
return nucleobase.lower()=='t'
def isPurine(nucleobase):
""" True if adenine (A) or guanine (G)
"""
return isAdenine(nucleobase) or isGuanine(nucleobase)
def isPyrimidine(nucleobase):
""" True if cytosine (C) or thymine (T)
"""
return isCytosine(nucleobase) or isThymine(nucleobase)
def isEqual(nucleobase1, nucleobase2):
""" Equal ignore case
"""
return nucleobase1.lower()==nucleobase2.lower()
通常,如果您有多個條件,則可以先將它們分配給變量。 可能不是您要找的答案,但可能可以幫助您編寫易於閱讀的代碼。
像這樣:
letter1 = "GT"
letter2 = "AC"
def transition(letter1, letter2):
cond1 = (letter1.lower() == 'gt')
cond2 = (letter2.lower() == 'ac')
cond3 = (letter1.lower() == 'ac')
cond4 = (letter2.lower() == 'gt')
if (cond1 and cond2) or (cond3 and cond4):
return True
else:
return False
transition(letter1,letter2)
嘗試這個:
def transition(letter1, letter2):
"""
>>> transition('G', 'A')
True
>>> transition('t', 'g')
False
>>> transition('C', 'c')
False
"""
if letter1.lower() == letter2.lower():
return False
return (letter1.lower() in 'ag' and letter2.lower() in 'ag') or (letter1.lower() in 'ct' and letter2.lower() in 'ct')
和
def transversion(letter1, letter2):
"""
>>> transversion('G', 'A')
False
>>> transition('t', 'g')
True
>>> transition('C', 'c')
False
"""
if letter1.lower() == letter2.lower():
return False
return (letter1.lower() in 'ac' and letter2.lower() in 'ac') or (letter1.lower() in 'gt' and letter2.lower() in 'gt')
似乎與letter1.lower() == 'gt'
您正在嘗試檢查字母是g還是t。 您可以使用in
代替==
。
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