使用pandas比较两列

Question

以此为出发点:

a = [['10', '1.2', '4.2'], ['15', '70', '0.03'], ['8', '5', '0']]
df = pd.DataFrame(a, columns=['one', 'two', 'three'])

Out[8]: 
  one  two three
0   10  1.2   4.2
1   15  70   0.03
2    8   5     0

我想if在熊猫中使用类似声明的东西.

if df['one'] >= df['two'] and df['one'] <= df['three']:
    df['que'] = df['one']

基本上,通过if语句检查每一行,创建新列.

文档说要使用,.all但没有例子......

Answer 1

你可以使用np.where.如果cond是布尔数组,A并且B是数组,那么

C = np.where(cond, A, B)

将C定义为Awhere 等于condTrue,B其中condFalse.

import numpy as np
import pandas as pd

a = [['10', '1.2', '4.2'], ['15', '70', '0.03'], ['8', '5', '0']]
df = pd.DataFrame(a, columns=['one', 'two', 'three'])

df['que'] = np.where((df['one'] >= df['two']) & (df['one'] <= df['three'])
                     , df['one'], np.nan)

产量

  one  two three  que
0  10  1.2   4.2   10
1  15   70  0.03  NaN
2   8    5     0  NaN

如果您有多个条件,则可以使用np.select.例如,如果你想df['que']等于df['two']什么时候df['one'] < df['two'],那么

conditions = [
    (df['one'] >= df['two']) & (df['one'] <= df['three']), 
    df['one'] < df['two']]

choices = [df['one'], df['two']]

df['que'] = np.select(conditions, choices, default=np.nan)

产量

  one  two three  que
0  10  1.2   4.2   10
1  15   70  0.03   70
2   8    5     0  NaN

如果我们可以假设什么df['one'] >= df['two']时候df['one'] < df['two']是假的,那么条件和选择可以简化为

conditions = [
    df['one'] < df['two'],
    df['one'] <= df['three']]

choices = [df['two'], df['one']]

(如果df['one']或df['two']包含NaN,则假设可能不正确.)

注意

a = [['10', '1.2', '4.2'], ['15', '70', '0.03'], ['8', '5', '0']]
df = pd.DataFrame(a, columns=['one', 'two', 'three'])

使用字符串值定义DataFrame.由于它们看起来是数字的,因此最好将这些字符串转换为浮点数:

df2 = df.astype(float)

然而,这会改变结果,因为字符串逐字符比较,而浮点数在数字上进行比较.

In [61]: '10' <= '4.2'
Out[61]: True

In [62]: 10 <= 4.2
Out[62]: False

Answer 2

您可以使用.equals列或整个数据帧.

df['col1'].equals(df['col2'])

如果它们相等,则该语句将返回True,否则False.

Answer 3

您可以使用apply()并执行类似的操作

df['que'] = df.apply(lambda x : x['one'] if x['one'] >= x['two'] and x['one'] <= x['three'] else "", axis=1)

或者如果你不想使用lambda

def que(x):
    if x['one'] >= x['two'] and x['one'] <= x['three']:
        return x['one']
    else:
        ''
df['que'] = df.apply(que, axis=1)

Answer 4

将每个单独的条件括在括号中，然后使用&运算符组合条件：

df.loc[(df['one'] >= df['two']) & (df['one'] <= df['three']), 'que'] = df['one']

您可以通过仅使用~（“not”运算符）来反转匹配来填充不匹配的行：

df.loc[~ ((df['one'] >= df['two']) & (df['one'] <= df['three'])), 'que'] = ''

您需要使用&and~而不是andandnot因为&and~运算符逐个元素地工作。

最终结果：

df
Out[8]: 
  one  two three que
0  10  1.2   4.2  10
1  15   70  0.03    
2   8    5     0  

Answer 5

一种方法是使用布尔系列来索引列df['one'].这会为您提供一个新列,其中的True条目具有与该行相同的值,df['one']并且False值为NaN.

布尔系列由您的if语句给出(尽管有必要使用&而不是and):

>>> df['que'] = df['one'][(df['one'] >= df['two']) & (df['one'] <= df['three'])]
>>> df
    one two three   que
0   10  1.2 4.2      10
1   15  70  0.03    NaN
2   8   5   0       NaN

如果希望将NaN值替换为其他值,则可以fillna在新列上使用该方法que.我在这里用0而不是空字符串:

>>> df['que'] = df['que'].fillna(0)
>>> df
    one two three   que
0   10  1.2   4.2    10
1   15   70  0.03     0
2    8    5     0     0

Answer 6

我想为那些尝试比较具有值的两列中的值的相等性NaN并False在两个值均为 时获取的人添加此答案NaN。根据定义，NaN!= NaN（参见：numpy.isnan(value) 与 value == numpy.nan 不同？）。

如果你想让两个NaN比较返回True，你可以使用：

df['compare'] = (df["col_1"] == df["col_2"]) | (df["col_1"].isna() & df["col_2"].isna())