Question

我有一个包含两列的文件，如下所示

-34G     trendmar
+41G     trendmar
 1.4G    ttpsenet
 3.6G    tyibahco
-13M     uberhelp
+19M     uberhelp
-8.9G    umljgate
+9.2G    umljgate

我想将它存储在字典中以进行一些数学运算，但是使用第一列作为值，第二列作为键。

我怎么能这样做？

Answer 1

您可以逐行读取文件，在空格上拆分并反向使用元素来创建字典：

with open("your_file", "r") as f:  # open the file for reading
    # read it line by line, split and invert the fields to use as k:v for your dict
    data = dict(reversed(line.split()) for line in f)
    # {'trendmar': '+41G', 'ttpsenet': '1.4G', 'tyibahco': '3.6G',
    #  'uberhelp': '+19M', 'umljgate': '+9.2G'}

请注意dict本质上是一个哈希映射，因此它不能有重复的密钥 - 如果重复密钥的值出现在文件中，它们将被最新值覆盖。

更新：如果您想保留所有值，您必须将它们存储为列表，例如：

import collections

data = collections.defaultdict(list)  # initiate all fields as lists
with open("your_file", "r") as f:  # open the file for reading
    for line in f:  # read the file line by line
        value, key = line.split()  # split the line to value and key
        data[key].append(value)  # append the value to the list for its key

现在你的data看起来像是：

{'trendmar': ['-34G', '+41G'], 'ttpsenet': ['1.4G'], 'tyibahco': ['3.6G'],
 'uberhelp': ['-13M', '+19M'], 'umljgate': ['-8.9G', '+9.2G']}

更新2：如果您想总结，而不是你＆＃39的值;会首先需要把它们转换成浮筒，然后用普通的算术运算，以达到最终值，所以先写一个函数将SI简写表示法转换为原始float：

QUANTIFIER_MAP = {"p": 1e15, "t": 1e12, "g": 1e9, "m": 1e6, "k": 1e3}
def si_to_float(number):
    try:
        last_char = number[-1].lower()
        if last_char in QUANTIFIER_MAP:
            return float(number[:-1]) * QUANTIFIER_MAP[last_char]
        return float(number)
    except ValueError:
        return 0.0

现在，您可以在创建list时将float替换为data并将值相加而不是追加：

import collections

data = collections.defaultdict(float)  # initiate all fields as integers
with open("your_file", "r") as f:  # open the file for reading
    # read it line by line, split and invert the fields to use as k:v for your dict
    for line in f:  # read the file line by line
        value, key = line.split()  # split the line to value and key
        data[key] += si_to_float(value)  # convert to float and add to the value for this key

这将导致data为：

{'trendmar': 7000000000.0, 'ttpsenet': 1400000000.0, 'tyibahco': 3600000000.0,
 'uberhelp': 6000000.0, 'umljgate': 300000000.0}

如果您想将这些值返回到SI缩短的表示法中，您必须编写si_to_float()的相反功能，然后转换所有data使用它的值，即：

QUANTIFIER_STACK = ((1e15, "p"), (1e12, "t"), (1e9, "g"), (1e6, "m"), (1e3, "k"))
def float_to_si(number):
    for q in QUANTIFIER_STACK:
        if number >= q[0]:
            return "{:.1f}".format(number / q[0]).rstrip("0").rstrip(".") + q[1].upper()
    return "{:.1f}".format(number).rstrip("0").rstrip(".")

# now lets traverse the previously created 'data' and convert its values:
for k, v in data.items():
    data[k] = float_to_si(v)

最终，这将导致data包含：

{'trendmar': '7G', 'ttpsenet': '1.4G', 'tyibahco': '3.6G',
 'uberhelp': '6M', 'umljgate': '300M'}

Answer 2

with open("file.txt","r") as file:
    print({e.split("     ")[1]:e.split("     ")[0] for e in file})

您可以使用词典理解

Answer 3

假设您希望将更多值与您的计算关键字相关联，这就是我的方法：

d = {}
with open("input.txt") as infile:
    lines = infile.readlines()
    keys = sorted(set(line.split()[1] for line in lines))
    for key in keys:
        tempList = []
        for line in lines:
            if line.split()[1]==key:
                tempList.append(line.split()[0])
        d.update({key:tempList})

print(d)

输出：

{'trendmar': ['-34G', '+41G'], 'uberhelp': ['-13M', '+19M'], 'umljgate': ['-8.9G', '+9.2G'], 'ttpsenet': ['1.4G'], 'tyibahco': ['3.6G']}

修改

如果您希望找到两个值之间的差异，可以使用literal_eval模块中的ast函数执行此操作，如下所示：

from ast import literal_eval d = {'trendmar': ['-34G', '+41G'], 'uberhelp': ['-13M', '+19M'], 'umljgate': ['-8.9G', '+9.2G'], 'ttpsenet': ['1.4G'], 'tyibahco': ['3.6G']} first = 0 second = 1 diff = [] for key in d.keys(): if len(d[key])==2: diff.append(key + " : " + str(literal_eval("".join([d[key][first][:-1] ," - (", d[key][second][:-1], ")"]))) + d[key][first][-1]) else: diff.append(key + " : " + str(literal_eval(str(d[key][0][:-1]))) + d[key][0][-1]) print(diff)

输出：

['uberhelp : -32M', 'tyibahco : 3.6G', 'ttpsenet : 1.4G', 'umljgate : -18.1G', 'trendmar : -75G']

在上面的例子中，我们从第二个值中减去第一个值。如果您希望相反，则交换first和second的值。

将文件输出存储在词典中

3 个答案: