04如何遍历pandas当中dataframe的元素 - 心默默言

04如何遍历pandas当中dataframe的元素

In [1]:

import pandas as pd

In [2]:

inp = [{\'c1\':10, \'c2\':100}, {\'c1\':11,\'c2\':110}, {\'c1\':12,\'c2\':120}]
df = pd.DataFrame(inp)
df

Out[2]:

 

	c1	c2
0	10	100
1	11	110
2	12	120

 

与此相关的有如下:

iterrows() : 将DataFrame迭代成（index ,series）

iteritems()： 将DataFrame迭代成（列名，series）

itertuples()： 将DataFrame迭代成元组

方法一：df.iterrows()

In [11]:

for index, row in df.iterrows():
    print(index, row["c1"], row["c2"])

 

0 10 100
1 11 110
2 12 120

In [14]:

for name, col in df.iteritems():
    print(name)
    print(col)

 

c1
0    10
1    11
2    12
Name: c1, dtype: int64
c2
0    100
1    110
2    120
Name: c2, dtype: int64

In [18]:

for i in df.itertuples():
    print(i)
    print(getattr(i, \'c1\'), getattr(i, \'c2\'))

 

Pandas(Index=0, c1=10, c2=100)
10 100
Pandas(Index=1, c1=11, c2=110)
11 110
Pandas(Index=2, c1=12, c2=120)
12 120

In [19]:

for i in df.itertuples(index=True, name=\'Pandas\'):
    print(i)
    print(getattr(i, \'c1\'), getattr(i, \'c2\'))

 

Pandas(Index=0, c1=10, c2=100)
10 100
Pandas(Index=1, c1=11, c2=110)
11 110
Pandas(Index=2, c1=12, c2=120)
12 120

In [20]:

for i in df.itertuples(index=False, name=\'nihao\'):
    print(i)
    print(getattr(i, \'c1\'), getattr(i, \'c2\'))

 

nihao(c1=10, c2=100)
10 100
nihao(c1=11, c2=110)
11 110
nihao(c1=12, c2=120)
12 120

 

第二种方案: applymap() 函数可以对DataFrame里的每个值进行处理,然后返回一个新的DataFrame

In [22]:

df = pd.DataFrame({
    \'a\': [1, 2, 3],
    \'b\': [10, 20, 30],
    \'c\': [5, 10, 15]
})
df

Out[22]:

 

	a	b	c
0	1	10	5
1	2	20	10
2	3	30	15

In [23]:

def add_one(x):
    return x + 1
df.applymap(add_one)

Out[23]:

 

	a	b	c
0	2	11	6
1	3	21	11
2	4	31	16

 

一个栗子:

这里有一组数据是10个学生的两次考试成绩,要求把成绩转换成ABCD等级:</br>

转换规则是:</br>

90-100 -> A</br> 80-89 -> B</br> 70-79 -> C</br> 60-69 -> D</br> 0-59 -> F</br>

In [24]:

grades_df = pd.DataFrame(
    data={\'exam1\': [43, 81, 78, 75, 89, 70, 91, 65, 98, 87],
          \'exam2\': [24, 63, 56, 56, 67, 51, 79, 46, 72, 60]},
    index=[\'Andre\', \'Barry\', \'Chris\', \'Dan\', \'Emilio\',
           \'Fred\', \'Greta\', \'Humbert\', \'Ivan\', \'James\']
)
grades_df

Out[24]:

 

	exam1	exam2
Andre	43	24
Barry	81	63
Chris	78	56
Dan	75	56
Emilio	89	67
Fred	70	51
Greta	91	79
Humbert	65	46
Ivan	98	72
James	87	60

In [25]:

def convert_to_letter(score):
    if (score >= 90):
        return \'A\'
    elif (score >= 80):
        return \'B\'
    elif (score >= 70):
        return \'C\'
    elif (score >= 60):
        return \'D\'
    else:
        return \'F\'
grades_df.applymap(convert_to_letter)

Out[25]:

 

	exam1	exam2
Andre	F	F
Barry	B	D
Chris	C	F
Dan	C	F
Emilio	B	D
Fred	C	F
Greta	A	C
Humbert	D	F
Ivan	A	C
James	B	D

 

第三种方案：iloc

In [26]:

df = pd.DataFrame({
    \'a\': [1, 2, 3],
    \'b\': [10, 20, 30],
    \'c\': [5, 10, 15]
})
df

Out[26]:

 

	a	b	c
0	1	10	5
1	2	20	10
2	3	30	15

In [28]:

for i in range(0, len(df)):
    print(df.iloc[i][\'a\'], df.iloc[i][\'b\'],df.iloc[i][\'c\'])

 

1 10 5
2 20 10
3 30 15