如何找到一個字符串的總和名單

Question

我有這樣

['MGM', '1'], ['MGD', '1'], ['V1', '[0,2,0,1]'], ['AuD', '[0,0,0,1]'] 

在python

列表。我想找到這樣的字符串列表的總和：

['MGM', '1'], ['MGD', '1'], ['V1', '3'], ['AuD', '1'] 

我應該將它們首先轉換爲列表中的列表嗎？如果是的話，我會怎麼做呢？

努力學習如何編寫科學代碼。

Answer 1

考慮：

s = [['MGM', '1'], ['MGD', '1'], ['V1', '[0,2,0,1]'], ['AuD', '[0,0,0,1]']] 

下面將第二項轉換爲Python對象：

import ast 
for sublist in s: 
    sublist[1] = ast.literal_eval(sublist[1]) 

結果：

[['MGM', 1], ['MGD', 1], ['V1', [0, 2, 0, 1]], ['AuD', [0, 0, 0, 1]]] 

然後將它們轉換回來的列表進行特殊處理：

for sublist in s: 
    if isinstance(sublist[1],list): 
     sublist[1] = sum(sublist[1]) 
    sublist[1] = str(sublist[1]) 

結果：

[['MGM', '1'], ['MGD', '1'], ['V1', '3'], ['AuD', '1']] 

Answer 2

這應該做你想要什麼。解釋在評論中。

import ast 

# Create the list 
myList = [['MGM', '1'], ['MGD', '1'], ['V1', '[0,2,0,1]'], ['AuD', '[0,0,0,1]']] 

# Loop through each sublist 
for num in myList: 
    # try to convert the string to a list and sum it 
    try: 
     # This works by evaluating the string into a list object 
     # Then summing the numbers in the list 
     # then turning that number back into a string so it's like the rest 
     num[1] = str(sum(ast.literal_eval(num[1]))) 
    # If it fails, it must just be a number, so ignore 
    except TypeError: 
     pass 

print myList 

Answer 3

假設你有一個字符串列表的列表，你可以在一個行做到這一點，利用ast.literal_eval()（評估字符串作爲一個列表），然後使用它sum()。示例 -

lst1 = [[a, (str(sum(ast.literal_eval(b))) if b.startswith('[') else b)] for a,b in lst] 

儘管它有點難以理解。更多使用for循環閱讀的版本有一點 -

lst1 = [] 
for a,b in lst: 
    if b.startswith('['): 
     lst1.append([a, (str(sum(ast.literal_eval(b)))]) 
    else: 
     lst1.append([a,b]) 

演示 -

>>> lst = [['MGM', '1'], ['MGD', '1'], ['V1', '[0,2,0,1]'], ['AuD', '[0,0,0,1]']] 
>>> import ast 
>>> lst1 = [[a, (str(sum(ast.literal_eval(b))) if b.startswith('[') else b)] for a,b in lst] 
>>> lst1 
[['MGM', '1'], ['MGD', '1'], ['V1', '3'], ['AuD', '1']] 

Answer 4

代碼：

from ast import literal_eval 
los=[['MGM', '1'], ['MGD', '1'], ['V1', '[0,2,0,1]'], ['AuD', '[0,0,0,1]']] 
for lst in los : 
    if '[' in lst[1]: 
     lst[1]=sum(literal_eval(lst[1])) 
print los 

Answer 5

>>> s = [['MGM', '1'], ['MGD', '1'], ['V1', '[0,2,0,1]'], ['AuD', '[0,0,0,1]']] 

因爲你的數據是一個災難，我們將使用一個幫手函數爲總和

>>> def mysum(L): 
...  try: 
...   return sum(L) 
...  except TypeError: 
...   return L 
... 

可以使用JSON或literal_eval琴絃

>>> import json 
>>> [[i, str(mysum(json.loads(j)))] for i, j in s] 
[['MGM', '1'], ['MGD', '1'], ['V1', '3'], ['AuD', '1']] 

Answer 6

感謝馬克，沒有人提到eval解碼！

lst = [['MGM', '1'], ['MGD', '1'], ['V1', '[0,2,0,1]'], ['AuD', '[0,0,0,1]']] 
for sublist in lst: 
    sublist[1] = eval(sublist[1]) 
for sublist in lst: 
    if isinstance(sublist[1],list): 
     sublist[1] = sum(sublist[1]) 
    sublist[1] = str(sublist[1]) 
print lst 

另外一個可以爲─

lst = [['MGM', '1'], ['MGD', '1'], ['V1', '[0,2,0,1]'], ['AuD', '[0,0,0,1]']] 

for x in lst: 
    x[1] = str(sum([int(i) for i in x[1].replace("[","").replace("]","").split(',')])) 

print lst 

它打印

[['MGM', '1'], ['MGD', '1'], ['V1', '3'], ['AuD', '1']]