Question

我有这个文本文件：

English 
hello bye 
italian 
ciao hola
spanish
hola chao

我想从每2个连续的行创建一个字典：

{
    'English': 'hello bye',
    'italian': 'ciao hola',
    'spanish': 'hola chao',
}

这是我的代码：

d= {}
with open("test.txt", 'r') as f:
    l = f.readlines()
    for line in l:
        (key,val) = line.split()
        d[key]=val

我得到了错误：

太多值无法解压缩错误

Answer 1

i = 0
d = {}
prev_key = None
for line in l:
    if i % 2 == 0:
        prev_key = line
    else:
        d[prev_key] = line
    i += 1

Answer 2

您也可以使用这种方法：

d = {}
with open("test.txt", 'r') as f:
    l = f.readlines()
    i = 0
    while i < len(l):
        d[l[i].replace("\n","")] = l[i+1].replace("\n","")
        i += 2

Answer 3

在原始代码中，您使用f.readlines()一次读取文件中的所有行，然后拆分每一行。问题在于，并不是每行都给您一个包含两个元素的列表，所以key, val = line.split()给您一个values to unpack，因为您试图将单个元素列表分配给两个项目。例如a,b = [2]会导致这种错误。

In [66]: a,b = [2]                                                                                                                                            
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-66-f9f79b7d1d3c> in <module>
----> 1 a,b = [2]

ValueError: not enough values to unpack (expected 2, got 1)

为避免这种情况，我们仅遍历读取的行，每个偶数元素都是键，而每个奇数元素都是字典中的值。

dct = {}
with open("file.txt", 'r') as f:
    l = f.readlines()
    idx = 0
    while idx < len(l):
        #Even element is key, Odd element is value
        key = l[idx].strip()
        value = l[idx+1].strip()
        dct[key] = value
        idx+=2
#{'English': 'hello bye', 'italian': 'ciao hola', 'spanish': 'hola chao'}

或更简单的使用dict-comprehension解决方案是

l = []
with open("file.txt", 'r') as f:
    l = f.readlines()

#This will be a list of tuples, with the first element of tuple being the key #and second value being the value
#Keys are obtained by slicing all even indexes, and values by slicing all odd indexes
key_value_tups = zip(l[::2], l[1::2])
#[('English \n', 'hello bye \n'), ('italian \n', 'ciao hola\n'), ('spanish\n', 'hola chao\n')]

#Iterate through the tuples and create the dict via dict-comprehension
dct = {key.strip() : value.strip() for key, value in key_value_tups}

print(dct)
#{'English': 'hello bye', 'italian': 'ciao hola', 'spanish': 'hola chao'}

Answer 4

您可以在一行中完成它：

with open("test.txt", 'r') as f:
    lines = f.readlines()

dict( zip( lines[::2], lines[1::2] ) )

lines[::2]将为您提供lines的所有具有偶数索引的元素
lines[1::2]将为您提供lines的所有具有奇数索引的元素 zip将根据两个列表创建一个迭代器(list1 elem, list2 elem)
dict将迭代器中的每个元组(key, value)用作字典项并创建字典

那一行相当于：

keys = []
values = []
for index, elem in enumerate(lines):
    if index % 2 == 0:
        keys += [elem]
    else:
        values += [elem]

d = {}
for key, val in zip(keys, values):
    d[key] = val

Answer 5

通过zip()使用字典理解：

with open("test.txt", 'r') as f:
    l = f.readlines()
    d = {x: y for x, y in zip(l[::2], l[1::2])}

文本文件中的字典

5 个答案: