Question

首先，我还是Python新手并且已经搜索过并且无法在任何地方找到如何做到这一点（从一个新人的角度来看）...

我有一个python

DataFrame

我需要打印出索引，列名和值。

我们说我有以下数据框

EAT     DAILY  WEEKLY  YEARLY
Fruit                        
APPLE       2       5     200
ORANGE      1       3     100
BANANA      1       4     150
PEAR        0       1      40

我需要打印它，以便我得到类似下面的内容，以便迭代数据帧中的每一行。

Eat Apple Daily at least 2
Eat Apple Weekly at least 5
Eat Apple Yearly at least 200
Eat Orange Daily at least 1
Eat Orange Weekly at least 3
Eat Orange Yearly at least 100
..
...
....

我尝试了各种组合但仍在学习，所以任何帮助都表示赞赏。

到目前为止，我已经尝试了

for row in test.iterrows():
    index, data = row
    print index , (data['column1'])
    print index , (data['column2'])
    print index , (data['column3'])

这将给我索引和值，但不是列加上我希望能够迭代，无论使用了多少列或行。另外，我仍然需要能够插入需要动态的文本......

Answer 1

您可以使用stack重塑Series MultiIndex，然后使用format Series.iteritems进行迭代：

test = test.stack()
print (test)
Fruit   EAT   
APPLE   DAILY       2
        WEEKLY      5
        YEARLY    200
ORANGE  DAILY       1
        WEEKLY      3
        YEARLY    100
BANANA  DAILY       1
        WEEKLY      4
        YEARLY    150
PEAR    DAILY       0
        WEEKLY      1
        YEARLY     40
dtype: int64

for index, data in test.iteritems():
    print (('Eat {} {} at least {}').format(index[0], index[1], data))

Eat APPLE DAILY at least 2
Eat APPLE WEEKLY at least 5
Eat APPLE YEARLY at least 200
Eat ORANGE DAILY at least 1
Eat ORANGE WEEKLY at least 3
Eat ORANGE YEARLY at least 100
Eat BANANA DAILY at least 1
Eat BANANA WEEKLY at least 4
Eat BANANA YEARLY at least 150
Eat PEAR DAILY at least 0
Eat PEAR WEEKLY at least 1
Eat PEAR YEARLY at least 40

但如果确实需要DataFrame添加reset_index然后按DataFrame.iterrows循环：

test = test.stack().reset_index(name='VAL')
print (test)
     Fruit     EAT  VAL
0    APPLE   DAILY    2
1    APPLE  WEEKLY    5
2    APPLE  YEARLY  200
3   ORANGE   DAILY    1
4   ORANGE  WEEKLY    3
5   ORANGE  YEARLY  100
6   BANANA   DAILY    1
7   BANANA  WEEKLY    4
8   BANANA  YEARLY  150
9     PEAR   DAILY    0
10    PEAR  WEEKLY    1
11    PEAR  YEARLY   40

for index, data in test.iterrows():
    print (('Eat {} {} at least {}').format(data['Fruit'], data['EAT'], data['VAL']))

Eat APPLE DAILY at least 2
Eat APPLE WEEKLY at least 5
Eat APPLE YEARLY at least 200
Eat ORANGE DAILY at least 1
Eat ORANGE WEEKLY at least 3
Eat ORANGE YEARLY at least 100
Eat BANANA DAILY at least 1
Eat BANANA WEEKLY at least 4
Eat BANANA YEARLY at least 150
Eat PEAR DAILY at least 0
Eat PEAR WEEKLY at least 1
Eat PEAR YEARLY at least 40

Answer 2

字符串系列

f = 'Eat {Fruit} {EAT} at least {value}'.format
df.stack().reset_index(name='value').apply(lambda x: f(**x), 1)

0         Eat APPLE DAILY at least 2
1        Eat APPLE WEEKLY at least 5
2      Eat APPLE YEARLY at least 200
3        Eat ORANGE DAILY at least 1
4       Eat ORANGE WEEKLY at least 3
5     Eat ORANGE YEARLY at least 100
6        Eat BANANA DAILY at least 1
7       Eat BANANA WEEKLY at least 4
8     Eat BANANA YEARLY at least 150
9          Eat PEAR DAILY at least 0
10        Eat PEAR WEEKLY at least 1
11       Eat PEAR YEARLY at least 40
dtype: object

打印

for idx, value in df.stack().iteritems():
    print('Eat {0[0]} {0[1]} at least {1}'.format(idx, value))

Eat APPLE DAILY at least 2
Eat APPLE WEEKLY at least 5
Eat APPLE YEARLY at least 200
Eat ORANGE DAILY at least 1
Eat ORANGE WEEKLY at least 3
Eat ORANGE YEARLY at least 100
Eat BANANA DAILY at least 1
Eat BANANA WEEKLY at least 4
Eat BANANA YEARLY at least 150
Eat PEAR DAILY at least 0
Eat PEAR WEEKLY at least 1
Eat PEAR YEARLY at least 40

Answer 3

甚至考虑使用pandas.DataFrame.to_string的非循环解决方案：

sdf = df.stack().reset_index(name='VALUE')
sdf['Output'] = sdf.apply(lambda row: "EAT {} {} at least {}".\
                          format(row['Fruit'], row['EAT'], row['VALUE']), axis=1)

# PRINT TO CONSOLE
print(sdf[['Output']].to_string(header=False, index=False, justify='left'))

# WRITE TO TEXT
with open('Output.txt', 'w') as f:
    f.write(sdf[['Output']].to_string(header=False, index=False, justify='left'))

# EAT APPLE DAILY at least 2
#    EAT APPLE WEEKLY at least 5
#  EAT APPLE YEARLY at least 200
#    EAT ORANGE DAILY at least 1
#   EAT ORANGE WEEKLY at least 3
# EAT ORANGE YEARLY at least 100
#    EAT BANANA DAILY at least 1
#   EAT BANANA WEEKLY at least 4
# EAT BANANA YEARLY at least 150
#      EAT PEAR DAILY at least 0
#     EAT PEAR WEEKLY at least 1
#    EAT PEAR YEARLY at least 40

您会注意到该方法目前存在reported bug的理由问题。当然，您可以通过字符串处理（strip()，replace()）进行补救，基本Python。

迭代数据帧以打印索引和列以及值

3 个答案: