我正在尝试创建一个名为“common_ancestor()”的函数,它接受两个输入:第一个是字符串分类名称列表,第二个是系统发育树字典。它应该返回一个字符串,给出所有分类中最接近共同祖先的分类单元的名称 物种在输入清单中。已经创建了一个名为“list_ancestors”的独立函数,它为我提供了列表中元素的一般祖先。另外,有一本我正在使用的词典。
tax_dict = {
'Pan troglodytes': 'Hominoidea', 'Pongo abelii': 'Hominoidea',
'Hominoidea': 'Simiiformes', 'Simiiformes': 'Haplorrhini',
'Tarsius tarsier': 'Tarsiiformes', 'Haplorrhini': 'Primates',
'Tarsiiformes': 'Haplorrhini', 'Loris tardigradus':'Lorisidae',
'Lorisidae': 'Strepsirrhini', 'Strepsirrhini': 'Primates',
'Allocebus trichotis': 'Lemuriformes', 'Lemuriformes': 'Strepsirrhini',
'Galago alleni': 'Lorisiformes', 'Lorisiformes': 'Strepsirrhini',
'Galago moholi': 'Lorisiformes'
}
def halfroot(tree):
taxon = random.choice(list(tree))
result = [taxon]
for i in range(0,len(tree)):
result.append(tree.get(taxon))
taxon = tree.get(taxon)
return result
def root(tree):
rootlist = halfroot(tree)
rootlist2 = rootlist[::-1]
newlist = []
for e in range(0,len(rootlist)):
if rootlist2[e] != None:
newlist.append(rootlist2[e])
return newlist[0]
def list_ancestors(taxon, tree):
result = [taxon]
while taxon != root(tree):
result.append(tree.get(taxon))
taxon = tree.get(taxon)
return result
def common_ancestors(inputlist,tree)
biglist1 = []
for i in range(0,len(listname)):
biglist1.append(list_ancestors(listname[i],tree))
"continue so that I get three separate lists where i can cross reference all elements from the first list to every other list to find a common ancestor "
结果应该类似于
print(common_ancestor([’Hominoidea’, ’Pan troglodytes’,’Lorisiformes’], tax_dict)
Output: ’Primates’"
答案 0 :(得分:0)
一种方法是收集每个物种的所有祖先,将它们放在一个集合中然后得到一个交叉点以获得它们的共同点:
def common_ancestor(species_list, tree):
result = None # initiate a `None` result
for species in species_list: # loop through each species in the species_list
ancestors = {species} # initiate the ancestors set with the species itself
while True: # rinse & repeat until there are leaves in the ancestral tree
try:
species = tree[species] # get the species' ancestor
ancestors.add(species) # store it in the ancestors set
except KeyError:
break
# initiate the result or intersect it with ancestors from the previous species
result = ancestors if result is None else result & ancestors
# finally, return the ancestor if there is only one in the result, or None
return result.pop() if result and len(result) == 1 else None
print(common_ancestor(["Hominoidea", "Pan troglodytes", "Lorisiformes"], tax_dict))
# Primates
您可以使用'中间'也是list_ancestors()
的这个函数的一部分 - 没有必要通过试图找到树的根来使它复杂化:
def list_ancestors(species, tree, include_self=True):
ancestors = [species] if include_self else []
while True:
try:
species = tree[species]
ancestors.append(species)
except KeyError:
break
return ancestors
当然,两者都依赖于一个有效的祖先树词典 - 如果一些祖先要自己递归,或者如果链条出现断裂则它不会起作用。此外,如果你要做很多这些操作,将平面字典变成合适的树可能是值得的。