相对于python中列表的col2中的值，在col1中查找最大值答案

【问题标题】：Finding max value in col1 with respect to values in col2 of a list in python相对于python中列表的col2中的值，在col1中查找最大值
【发布时间】：2018-06-13 07:17:19
【问题描述】：

我是 python 新手。我想从 col2 中找到与列表 col1 中的值 'men'、'women' 和 'people' 相关的最大值。比如，['men', 12, '1946-Truman.txt'], ['women', 7, '1946-Truman.txt']and['people', 49, '1946-Truman.txt'] 包含男性、女性和人的 col2 的最大值。

一种可能的解决方案是将此元组列表转换为男性、女性和人的三个单独数组，然后从所有数组中找到最大值。但是，我想要一个更好的解决方案。

数据：

[['men', 2, '1945-Truman.txt']
['women', 2, '1945-Truman.txt']
['people', 10, '1945-Truman.txt']
['men', 12, '1946-Truman.txt']
['women', 7, '1946-Truman.txt']
['people', 49, '1946-Truman.txt']
['men', 7, '1947-Truman.txt']
['women', 2, '1947-Truman.txt']
['people', 12, '1947-Truman.txt']
['men', 4, '1948-Truman.txt']
['women', 1, '1948-Truman.txt']
['people', 22, '1948-Truman.txt']
['men', 2, '1949-Truman.txt']
['women', 1, '1949-Truman.txt']
['people', 15, '1949-Truman.txt']
['men', 6, '1950-Truman.txt']
['women', 2, '1950-Truman.txt']
['people', 15, '1950-Truman.txt']
['men', 8, '1951-Truman.txt']
['women', 2, '1951-Truman.txt']
['people', 9, '1951-Truman.txt']
['men', 3, '1953-Eisenhower.txt']
['women', 0, '1953-Eisenhower.txt']
['people', 17, '1953-Eisenhower.txt']]

提前致谢。

【问题讨论】：

标签： python list numpy tuples max

【解决方案1】：

pandas 很好，但你可以使用max 和lambda：

men = max(data, key=lambda x: x[1] if x[0] == 'men' else 0)
women = max(data, key=lambda x: x[1] if x[0] == 'women' else 0)
people = max(data, key=lambda x: x[1] if x[0] == 'people' else 0)

【讨论】：

谢谢。太好了。

【解决方案2】：

您可以使用pandas 包。通过定义数据框：

import pandas as pd
df = pd.DataFrame([['men', 2, '1945-Truman.txt'],
                   ['women', 2, '1945-Truman.txt'],
                   ['people', 10, '1945-Truman.txt'],
                   ['men', 12, '1946-Truman.txt'],
                    ['women', 7, '1946-Truman.txt'],
                   ['people', 49, '1946-Truman.txt'],
                   ['men', 7, '1947-Truman.txt'],
                   ['women', 2, '1947-Truman.txt'],
                   ['people', 12, '1947-Truman.txt'],
                   ['men', 4, '1948-Truman.txt'],
                   ['women', 1, '1948-Truman.txt'],
                   ['people', 22, '1948-Truman.txt'],
                   ['men', 2, '1949-Truman.txt'],
                   ['women', 1, '1949-Truman.txt'],
                   ['people', 15, '1949-Truman.txt'],
                   ['men', 6, '1950-Truman.txt'],
                   ['women', 2, '1950-Truman.txt'],
                   ['people', 15, '1950-Truman.txt'],
                   ['men', 8, '1951-Truman.txt'],
                   ['women', 2, '1951-Truman.txt'],
                   ['people', 9, '1951-Truman.txt'],
                   ['men', 3, '1953-Eisenhower.txt'],
                   ['women', 0, '1953-Eisenhower.txt'],
                   ['people', 17, '1953-Eisenhower.txt']])

然后

df.groupby([0], sort=False)[1].max()

返回

0 
men       12
women      7
people    49
Name: 1, dtype: int64

这是你想要的吗？

【讨论】：

【解决方案3】：

如果您使用的是列表列表，例如：

lst=[['men', 2123, '1945-Truman.txt'],
['women', 2, '1945-Truman.txt'],
['people', 10, '1945-Truman.txt'],
['men', 12, '1946-Truman.txt'],
['women', 7, '1946-Truman.txt'],
['people', 49, '1946-Truman.txt'],
['men', 7, '1947-Truman.txt'],
['women', 2, '1947-Truman.txt']]

然后你可以使用下面的代码。

max_men=0
max_women=0
max_people =0
for item in lst:
    if((item[0]=="men") and (item[1]>max_men)):
        max_men=item[1]
    elif((item[0]=="women") and (item[1]>max_women)):
        max_women=item[1]
    elif((item[0]=="people") and (item[1]>max_people)):
        max_people=item[1]

print max_men
print max_women
print max_people

这将进入名为lst 的位列表中的每个列表，并找到男性、女性和人的最大值。

【讨论】：

【解决方案4】：

您可以从第一列创建一个集合，然后找到最大值：

data = [
    ['men', 2, '1945-Truman.txt'],
    ['women', 2, '1945-Truman.txt'],
    ...
]

keys = set([col[0] for col in data])

for k in keys:
        print (k, max([col[1] for col in data if col[0] == k]))

返回：

women 7
people 49
men 12

【讨论】：

很好的答案。真的很容易理解。如果 .txt 文件的名称（col3）也打印在输出中，那就太好了。

【解决方案5】：

你可以使用itertools.groupby:

import itertools
new_data = [(a, list(b)) for a, b in itertools.groupby(sorted(data, key=lambda x:x[0]), key=lambda x:x[0])]
new_final_data = [max(b, key=lambda x:x[1]) for a, b in new_data]

输出：

[['men', 12, '1946-Truman.txt'], ['people', 49, '1946-Truman.txt'], ['women', 7, '1946-Truman.txt']]

或者，一个字典，每个键对应个人的类型：

new_final_data = {a:max(b, key=lambda x:x[1]) for a, b in new_data}

输出：

{'women': ['women', 7, '1946-Truman.txt'], 'men': ['men', 12, '1946-Truman.txt'], 'people': ['people', 49, '1946-Truman.txt']}

【讨论】：

【解决方案6】：

你可以使用pandas，我想data是一个列表列表：

import pandas as pd

df = pd.DataFrame(data)

df.loc[df.groupby([0])[1].idxmax()]

        0   1                2
3     men  12  1946-Truman.txt
5  people  49  1946-Truman.txt
4   women   7  1946-Truman.txt

对于相同格式的结果：

df.loc[df.groupby([0])[1].idxmax()].values.tolist()

[['men', 12, '1946-Truman.txt'], ['people', 49, '1946-Truman.txt'], ['women', 7, '1946-Truman.txt']]

【讨论】：

【解决方案7】：

men = [t for t in yourlist if t[0] == 'men']
women = [t for t in yourlist  if t[0] == 'women']
people = [t for t in yourlist  if t[0] == 'people']
sorted(men, key=operator.itemgetter(1), reverse=True)[0][1]
sorted(women, key=operator.itemgetter(1), reverse=True)[0][1]
sorted(people, key=operator.itemgetter(1), reverse=True)[0][1]

【讨论】：