遍历一个数据框来打印索引和列和值

首先，我对Python还是一个新手，并且已经搜索并且无法找到任何地方如何做到这一点（从一个新的角度来看）...遍历一个数据框来打印索引和列和值

我有一个python

DataFrame

我需要打印索引，列名和值。

比方说，我有以下的数据帧

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

我需要打印我们这样的，我会得到像下面这样它的每一行遍历的数据帧。

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，但我仍然在学习，所以任何帮助表示赞赏。

到目前为止，我已经尝试

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

哪位能给我的指数和值，但不会列再加上我希望它能够不管迭代多少列或行使用。另外，我还需要能够插入其需要是动态的文字...

来源

2017-02-20 maddhatter1219

系列串

的

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

打印出来

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

来源

2017-02-20 06:29:05 piRSquared

我曾经遇到过堆栈和interitems，但不确定让我到哪里需要的语法。完全按照我需要的方式工作。 – maddhatter1219

可以使用stack为重塑到Series与MultiIndex，然后通过Series.iteritems与format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

但如果确实需要通过DataFrame添加DataFrame.iterrows和reset_index然后循环：

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

来源

2017-02-20 06:09:32 jezrael

考虑使用pandas.DataFrame.to_string甚至非循环方案：

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

你会发现一个理由问题目前是reported bug的方法。当然，你可以通过字符串处理（strip(),replace()）来解决一般情况下的基本Python问题。

来源

2017-02-20 15:44:56 Parfait

遍历一个数据框来打印索引和列和值

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