Python 如何找到两个列表之间的匹配项并根据匹配项写入输出？
我不确定我是否把问题的标题放得恰当。但是，我试着在下面解释这个问题。如果你能思考这个问题，请建议合适的标题 假设我有两种类型的列表数据：Python 如何找到两个列表之间的匹配项并根据匹配项写入输出？,python,python-2.7,for-loop,if-statement,list-comprehension,Python,Python 2.7,For Loop,If Statement,List Comprehension,我不确定我是否把问题的标题放得恰当。但是，我试着在下面解释这个问题。如果你能思考这个问题，请建议合适的标题 假设我有两种类型的列表数据： list_headers = ['gene_id', 'gene_name', 'trans_id'] # these are the features to be mined from each line of `attri_values` attri_values = ['gene_id "scaffold_200001.1"', 'gene_ve
list_headers = ['gene_id', 'gene_name', 'trans_id']
# these are the features to be mined from each line of `attri_values`
attri_values =
['gene_id "scaffold_200001.1"', 'gene_version "1"', 'gene_source "jgi"', 'gene_biotype "protein_coding"']
['gene_id "scaffold_200001.1"', 'gene_version "1"', 'trans_id "scaffold_200001.1"', 'transcript_version "1"', 'exon_number "1"', 'gene_source "jgi"', 'gene_biotype "protein_coding"', 'transcript_source "jgi"', 'transcript_biotype "protein_coding"', 'exon_id "scaffold_200001.1.exon1"', 'exon_version "1"']
['gene_id "scaffold_200002.1"', 'gene_version "1"', 'trans_id "scaffold_200002.1"', 'transcript_version "1"', 'exon_number "3"', 'gene_source "jgi"', 'gene_biotype "protein_coding"', 'transcript_source "jgi"', 'transcript_biotype "protein_coding"', 'exon_id "scaffold_200002.1.exon3"', 'exon_version "1"']
gene_id gene_name trans_id
scaffold_200001.1 NA NA
scaffold_200001.1 NA scaffold_200001.1
scaffold_200002.1 NA scaffold_200002.1
gene_id gene_name trans_id
scaffold_200001.1
scaffold_200001.1 scaffold_200001.1
scaffold_200002.1 scaffold_200002.1
列表和属性值中的属性的匹配情况创建一个表
output = open('gtf_table', 'w')
output.write('\t'.join(list_headers) + '\n') # this will first write the header
# then I want to read each line
for values in attri_values:
for list in list_headers:
if values.startswith(list):
attr_id = ''.join([x for x in attri_values if list in x])
attr_id = attr_id.replace('"', '').split(' ')[1]
output.write('\t' + '\t'.join([attr_id]))
elif not values.startswith(list):
attr_id = 'NA'
output.write('\t' + '\t'.join([attr_id]))
output.write('\n')
问题：是当在属性值的值中找到列表标题列表中的匹配字符串时，所有字符串都工作正常，但当没有匹配时，会有大量重复的“NA”
最终预期结果：
list_headers = ['gene_id', 'gene_name', 'trans_id']
# these are the features to be mined from each line of `attri_values`
attri_values =
['gene_id "scaffold_200001.1"', 'gene_version "1"', 'gene_source "jgi"', 'gene_biotype "protein_coding"']
['gene_id "scaffold_200001.1"', 'gene_version "1"', 'trans_id "scaffold_200001.1"', 'transcript_version "1"', 'exon_number "1"', 'gene_source "jgi"', 'gene_biotype "protein_coding"', 'transcript_source "jgi"', 'transcript_biotype "protein_coding"', 'exon_id "scaffold_200001.1.exon1"', 'exon_version "1"']
['gene_id "scaffold_200002.1"', 'gene_version "1"', 'trans_id "scaffold_200002.1"', 'transcript_version "1"', 'exon_number "3"', 'gene_source "jgi"', 'gene_biotype "protein_coding"', 'transcript_source "jgi"', 'transcript_biotype "protein_coding"', 'exon_id "scaffold_200002.1.exon3"', 'exon_version "1"']
gene_id gene_name trans_id
scaffold_200001.1 NA NA
scaffold_200001.1 NA scaffold_200001.1
scaffold_200002.1 NA scaffold_200002.1
gene_id gene_name trans_id
scaffold_200001.1
scaffold_200001.1 scaffold_200001.1
scaffold_200002.1 scaffold_200002.1
后期编辑：
这就是我如何编写elif
的问题（因为对于每一个不匹配项，它都会写入'NA'）。我尝试以不同的方式移动NA
的条件，但没有成功如果我删除elif
它会以（NA
丢失）的形式获得输出：
list_headers = ['gene_id', 'gene_name', 'trans_id']
# these are the features to be mined from each line of `attri_values`
attri_values =
['gene_id "scaffold_200001.1"', 'gene_version "1"', 'gene_source "jgi"', 'gene_biotype "protein_coding"']
['gene_id "scaffold_200001.1"', 'gene_version "1"', 'trans_id "scaffold_200001.1"', 'transcript_version "1"', 'exon_number "1"', 'gene_source "jgi"', 'gene_biotype "protein_coding"', 'transcript_source "jgi"', 'transcript_biotype "protein_coding"', 'exon_id "scaffold_200001.1.exon1"', 'exon_version "1"']
['gene_id "scaffold_200002.1"', 'gene_version "1"', 'trans_id "scaffold_200002.1"', 'transcript_version "1"', 'exon_number "3"', 'gene_source "jgi"', 'gene_biotype "protein_coding"', 'transcript_source "jgi"', 'transcript_biotype "protein_coding"', 'exon_id "scaffold_200002.1.exon3"', 'exon_version "1"']
gene_id gene_name trans_id
scaffold_200001.1 NA NA
scaffold_200001.1 NA scaffold_200001.1
scaffold_200002.1 NA scaffold_200002.1
gene_id gene_name trans_id
scaffold_200001.1
scaffold_200001.1 scaffold_200001.1
scaffold_200002.1 scaffold_200002.1
python为字符串提供了一个find
方法，您可以使用该方法为每个属性值迭代每个列表头。尝试使用此功能：
def Get_Match(search_space,search_string):
start_character = search_space.find(search_string)
if start_character == -1:
return "N/A"
else:
return search_space[(start_character + len(search_string)):]
for i in range(len(attri_values_1)):
for j in range(len(list_headers)):
print Get_Match(attri_values_1[i],list_headers[j])
python为字符串提供了一个find
方法，您可以使用该方法为每个属性值迭代每个列表头。尝试使用此功能：
def Get_Match(search_space,search_string):
start_character = search_space.find(search_string)
if start_character == -1:
return "N/A"
else:
return search_space[(start_character + len(search_string)):]
for i in range(len(attri_values_1)):
for j in range(len(list_headers)):
print Get_Match(attri_values_1[i],list_headers[j])
我的答案是使用熊猫
import pandas as pd
# input data
list_headers = ['gene_id', 'gene_name', 'trans_id']
attri_values = [
['gene_id "scaffold_200001.1"', 'gene_version "1"', 'gene_source "jgi"', 'gene_biotype "protein_coding"'],
['gene_id "scaffold_200001.1"', 'gene_version "1"', 'trans_id "scaffold_200001.1"', 'transcript_version "1"', 'exon_number "1"', 'gene_source "jgi"', 'gene_biotype "protein_coding"', 'transcript_source "jgi"', 'transcript_biotype "protein_coding"', 'exon_id "scaffold_200001.1.exon1"', 'exon_version "1"'],
['gene_id "scaffold_200002.1"', 'gene_version "1"', 'trans_id "scaffold_200002.1"', 'transcript_version "1"', 'exon_number "3"', 'gene_source "jgi"', 'gene_biotype "protein_coding"', 'transcript_source "jgi"', 'transcript_biotype "protein_coding"', 'exon_id "scaffold_200002.1.exon3"', 'exon_version "1"']]
# process input data
attri_values_X = [dict([tuple(b.split())[:2] for b in a]) for a in attri_values]
# Create DataFrame with the desired columns
df = pd.DataFrame(attri_values_X, columns=list_headers)
# print dataframe
print df
输出
gene_id gene_name trans_id
0 "scaffold_200001.1" NaN NaN
1 "scaffold_200001.1" NaN "scaffold_200001.1"
2 "scaffold_200002.1" NaN "scaffold_200002.1"
没有熊猫也很容易。我已经给了你attri\u values\u X
，然后你就快到了，只需从你不想要的字典中删除键。我的答案是使用pandas
import pandas as pd
# input data
list_headers = ['gene_id', 'gene_name', 'trans_id']
attri_values = [
['gene_id "scaffold_200001.1"', 'gene_version "1"', 'gene_source "jgi"', 'gene_biotype "protein_coding"'],
['gene_id "scaffold_200001.1"', 'gene_version "1"', 'trans_id "scaffold_200001.1"', 'transcript_version "1"', 'exon_number "1"', 'gene_source "jgi"', 'gene_biotype "protein_coding"', 'transcript_source "jgi"', 'transcript_biotype "protein_coding"', 'exon_id "scaffold_200001.1.exon1"', 'exon_version "1"'],
['gene_id "scaffold_200002.1"', 'gene_version "1"', 'trans_id "scaffold_200002.1"', 'transcript_version "1"', 'exon_number "3"', 'gene_source "jgi"', 'gene_biotype "protein_coding"', 'transcript_source "jgi"', 'transcript_biotype "protein_coding"', 'exon_id "scaffold_200002.1.exon3"', 'exon_version "1"']]
# process input data
attri_values_X = [dict([tuple(b.split())[:2] for b in a]) for a in attri_values]
# Create DataFrame with the desired columns
df = pd.DataFrame(attri_values_X, columns=list_headers)
# print dataframe
print df
输出
gene_id gene_name trans_id
0 "scaffold_200001.1" NaN NaN
1 "scaffold_200001.1" NaN "scaffold_200001.1"
2 "scaffold_200002.1" NaN "scaffold_200002.1"
没有熊猫也很容易。我已经给了你attri\u values\u X
，然后你就快到了，只需从你不想要的字典中删除键。我设法编写了一个函数，它将有助于解析你的数据。我试图修改您发布的原始代码，但使问题复杂的是您存储需要解析的数据的方式，无论如何我无法判断，以下是我的代码：
def searchHeader(title, values):
""""
searchHeader(title, values) --> list
*Return all the words of strings in an iterable object in which title is a substring,
without including title. Else write 'N\A' for strings that title is not a substring.
Example:
>>> seq = ['spam and ham', 'spam is awesome', 'Ham is...!', 'eat cake but not pizza']
>>> searchHeader('spam', attri_values)
['and', 'ham', 'is', 'awesome', 'N\\A', 'N\\A']
"""
res = []
for x in values:
if title in x:
res.append(x)
else:
res.append('N\A') # If no match found append N\A for every string in values
res = ' '.join(res)
# res = res.replace('"', '') You can use this for your code or use it after you call the function on res
res = res.split(' ')
res = [x for x in res if x != title] # Remove title string from res
return res
正则表达式在这种情况下也很方便。使用此函数解析数据，然后格式化结果以将表写入文件。此函数仅使用一个循环和一个列表理解，其中在代码中使用两个嵌套的循环和一个列表理解
将每个头字符串分别传递给函数，如下所示：
for title in list_headers:
result = searchHeader(title, attri_values)
...format as table...
...write to file...
如果可能的话，考虑从一个简单的列表移动到一个字典，用于<代码> ATTILIOVALUTION/CODE >，这样你就可以用它们的标题对字符串进行分组：
attri_values = {'header': ('data1', 'data2',...)}
在我看来，这比使用列表要好。还要注意，您正在代码中重写list
name，这不是一件好事，因为list
实际上是创建列表的内置类 我成功地编写了一个有助于解析数据的函数。我试图修改您发布的原始代码，但使问题复杂的是您存储需要解析的数据的方式，无论如何我无法判断，以下是我的代码：
def searchHeader(title, values):
""""
searchHeader(title, values) --> list
*Return all the words of strings in an iterable object in which title is a substring,
without including title. Else write 'N\A' for strings that title is not a substring.
Example:
>>> seq = ['spam and ham', 'spam is awesome', 'Ham is...!', 'eat cake but not pizza']
>>> searchHeader('spam', attri_values)
['and', 'ham', 'is', 'awesome', 'N\\A', 'N\\A']
"""
res = []
for x in values:
if title in x:
res.append(x)
else:
res.append('N\A') # If no match found append N\A for every string in values
res = ' '.join(res)
# res = res.replace('"', '') You can use this for your code or use it after you call the function on res
res = res.split(' ')
res = [x for x in res if x != title] # Remove title string from res
return res
正则表达式在这种情况下也很方便。使用此函数解析数据，然后格式化结果以将表写入文件。此函数仅使用一个循环和一个列表理解，其中在代码中使用两个嵌套的循环和一个列表理解
将每个头字符串分别传递给函数，如下所示：
for title in list_headers:
result = searchHeader(title, attri_values)
...format as table...
...write to file...
如果可能的话，考虑从一个简单的列表移动到一个字典，用于<代码> ATTILIOVALUTION/CODE >，这样你就可以用它们的标题对字符串进行分组：
attri_values = {'header': ('data1', 'data2',...)}
在我看来，这比使用列表要好。还要注意，您正在代码中重写list
name，这不是一件好事，因为list
实际上是创建列表的内置类 谢谢你的回答。使用字典会很复杂，因为这些只是大数据的一小部分。我认为简单的嵌套for循环可以解决这个问题。顺便说一句，我在
result=searchHeader（列出标题、属性值）`@everestial007我的坏！我应该将title
而不是list\u headers
传递给函数：result=searchHeader（title，attri\u value）
。这可能是深夜编写代码的后果吗：P？我理解太多地站在电脑前和/或昏昏欲睡的后果。顺便说一句，代码仍然不能为我解决这个问题。我试着改变一些东西，比如在x:
中改为if title:
我认为应该是if x.startswith（title）
原因是除非所有字符串都匹配，否则列表对列表的比较不会出现命中。我也尝试过改变其他事情，但没有运气。你能给我一个完整的工作例子吗那可能吗。请向上投票，让其他人也更加关注这个问题。@everestial007查看此代码以了解如何使用该函数，顺便说一句，我使用print
打印测试结果。您可以在将结果格式化为表后将其写入文件：。您只需将所提供函数的输出格式化为一个表，然后将其写入输出文件。@everestal007如果您一直在写表，请参见下面的。虽然它仍然为不匹配的字符串打印N/A，但我已经包含了N/A以满足zip
，zip
一直运行到其参数列表中最短的iterable，从而删除了其他元素；在我们的例子中，它从属性值中删除字符串值。请参阅zip
的文档以了解更多信息或运行帮助（zip）
。研究代码，你会发现它很容易，就像柠檬挤压一样！：）谢谢你的回答。使用字典会很复杂，因为这些只是大数据的一小部分。我认为简单的嵌套for循环可以解决这个问题。顺便说一句，我得到类型错误