如何根据 Python 列表中存在的值从列表中删除字典？

Question

我是 Python 的新手，如果我的代码没有以最 'pythonic' 的方式编写，请提前致歉。

我正在将 CSV 文件上传到脚本，如果该行符合特定条件，我想过滤该 CSV。

我有两个列表，a_lst 和 b_lst。一旦字典在 a_lst 中，我就会检查 b_lst 中是否有对应 key:value 的字典。如果有匹配项，则会将其打印到控制台。我不想打印到控制台，而是想从 a_lst 中删除该项目。我该怎么做？

a_lst = []
b_lsts = []

with open(file_name, 'rt') as f:
        reader = csv.DictReader(f)
        for row in reader:
            if row['Minutes'] == '0' and row['MB'] == '0' and row['Calls'] == '1':
                a_lst.append(row)
            elif row['Minutes'] == '0' and row['MB'] == '' and row['Calls'] == '1':
                a_lst.append(row)
            elif row['Minutes'] == '' and row['MB'] == '0' and row['Calls'] == '1':
                a_lst.append(row)
            elif row['Minutes'] == '' and row['MB'] == '' and row['Calls'] == '1':
                a_lst.append(row)
            else:
                b_lst.append(row)

i = 0
while i < len(a_lst):
    if not any(d['Name'] == a_lst[i]['Name'] for d in b_lst):
        print a_lst[i]['Name']+"(Row"+str(i)+") is not b_lst."
    else:
        print a_lst[i]['Name']+"(Row"+str(i)+") is present."
            i+=1

编辑：我想要的结果

Name, PhoneNo, Minutes, MB, Calls
Steve,0777777777,0,0,1
Steve,0777777777,0,2,14
Steve,0777777777,0,0,1
John,078888888,0,0,1
John,078888888,0,0,1
John,078888888,0,0,1
Dave,07999999,2,3,4
Dave,07999999,2,6,24

如果上面的数据是我插入的，我只想看到 John 的名字，因为他是唯一一个他的名字的所有行都包含值“0,0,1”的人'

Answer 1

如果元素具有相同的 key/value，只需从列表中删除该元素，您还想在 any 之前删除 not，因为如果有任何匹配，我们要删除：

for ele in a_lst[:]:
    if  any(d['Name'] == ele['Name'] for d in b_lst):
        a_lst.remove(ele)

或者在添加之前忘记使用 any 和过滤器，将 row['Name'] 添加到集合中并检查我们是否已经看到它：

seen = set()
with open(file_name, 'rt') as f:
        reader = csv.DictReader(f)
        for row in reader:
           if row['Name'] in seen:
               continue
           if all((row['Minutes'] == '0', (row['MB'] == '0' or not row['MB']), row['Calls'] == '1')):
               a_lst.append(row)
           elif all((not row['Minutes'], (row['MB'] or not row['MB']), row['Calls'] == '1')):
                a_lst.append(row)
           else:
               seen.add(row['Name']) 
         # remove "else:" and just use seen.add(row['Name']) outside the elif if you want all dups removed

根据您的编辑：

seen = set()
with open(infile, 'rt') as f:
    reader = csv.reader(f,delimiter=",")
    for row in reader:
        if row[0] in seen:
            continue
        if all(x in {"0", "1"} for x in row[2:]):
            print(row)
        seen.add(row[0])

输出：

['Steve', '0777777777', '0', '0', '1']
['John', '078888888', '0', '0', '1']

Steve 和 John 的相关列中只有 0 和 1。

如果您只想要列中仅包含 0 和 1 的名称：

from collections import defaultdict
d = defaultdict(list)

with open(infile, 'rt') as f:
    reader = csv.reader(f,delimiter=",")
    for row in reader:
        d[row[0]].append([row, set(row[2:])])

print([v[0][0] for k, v in d.items() if all(sub[1] == {"0","1"} for sub in v)])

[['John', '078888888', '0', '0', '1']]

或者如果您的名字总是组合在一起，则使用集合：

seen = set()
temp = set()

with open(infile, 'rt') as f:
    reader = csv.reader(f,delimiter=",")
    next(reader)
    prev = None
    for row in reader:
        # found new name and it is not the first
        if row[0] not in seen and temp:
            # set should only hav  and 1 if all columns only contain 0,1
            if temp == {"0", "1"}:
                print(prev)  # print previous row
            # reset temp
            temp = set()
        seen.add(row[0])
        temp.update(row[2:])
        # need to keep track of previous row 
        prev = row

输出：

['John', '078888888', '0', '0', '1']

如何根据 Python 列表中存在的值从列表中删除字典？

How do I remove a dictionary from a list depending on a value being present in a list in Python?

python

csv

dictionary

list