Question

我读了两个这样的.csv文件。

ori = "all.csv"
det = "find.csv"

names = []
namesa = []


with open(det, "r") as cursor:
    for row in cursor:
        cells = row.split(",")
        if len(cells) > 2:
            b = cells[1]
            c = b.split("-")
            names.append(c[0])


with open(ori, "r") as rcursor1: #read the document
    for trow in rcursor1: #read each row
        row1 = trow.split(",") #split it by your seperator
        namesa.append(row1)

工作得很好。 namesa是一个嵌套列表，其中.csv中的每一行都是一个列表（请参阅示例），而names包含我想在namesa中找到的值。如果names中的值在namesa中，我想要整个＆＃34;嵌套列表部分＆＃34;。那就是。

#example
namesa = [[a,b,c,], [a1, b1, c1], [xy, cd, e2], [u1, i1, il], ...]
names = [a, u1,]
return = [[a1, b1, c1], [u1, i1, il], ...] 

#or
namesa = [[john,bill,catherina,], [marti, alex, christoph], [ben, sherlock, london], [Bern, paris, Zürich], ...]
names = [sherlock, marti]
results = [[marti, alex, christoph],  [ben, sherlock, london]]

嗯，这不起作用。那就是我到目前为止所尝试的内容：

#did not return any match
d = list([b for b in namesa if b in [a for a in names]])
print d


#did not return any match neither
for a in namesa:
    for b in names:
        if b in a:
            print "match"


#well, that did not work neither
for a in namesa:
    for b in names:
        if a[5] == b:
            print "match"

没有比赛回来。我在excel中打开了两个csv文件并用手搜索了＃34;对于那些让我回复的比赛...... 我在这做错了什么？使用python。

Answer 1

namesa = [['john', 'bill', 'catherina'], ['cat', 'dog', 'foo'], ['noodle', 'bob']]
names = ['john','foo']

试试这个

for n in names:
    for arr in namesa:
        if n.strip() in ''.join(arr):
            print arr

.strip因为names列表中的值似乎有尾随空格。

Answer 2

如果您使用.csv文件，我建议您使用csv模块。我这样做（我假设你要找的东西是'姓'栏。如果它们在不同的栏目中，你可以考虑用它们进行迭代，或者做name in row['surname'] or name in row['name']，取决于并发症：

import csv
result = []
listFromCSV = []
names = ['alex','sherlock']
csvFile = open('yourFile.csv')
reader = csv.DictReader(csvFile)
fieldnames = reader.fieldnames
for row in reader:
    listFromCSV.append(row)
csvFile.close()
for name in names:
    for row in listFromCSV:
        if name.strip() in row['surname']:
            result.append(row)

如果你想摆脱重复，可以在最后一个句点附加break for循环。

Answer 3

namesa = [['john','bill','catherina',], ['marti', 'alex', 'christoph'], ['ben', 'sherlock', 'london']]

names = ['sherlock', 'marti']

for i in namesa:
  for j in names:
    if j in i:
        print i

输出

['marti'，'alex'，'christoph'] ['ben'，'sherlock'，'london']

从列表中查找嵌套列表中的匹配值

3 个答案: