pandas 数据框 - 返回 iloc 中的值，如果不存在则返回零

Question

In using the iloc method for Pandas dataframe, I want to return zero if the value does not exist: (I have a query which it will always return either one row or an empty dataframe. I want the first left value when it exists)

import pandas as pd

mydict = {"col1":[1,2], "price":[1000,2000]}
df = pd.DataFrame(mydict)
query=df[df['price']>3000]

try:
    print(query.iloc[0][0])
except BaseException:
    print(0)

#print result: 0

Is there any better way or built-in method for iloc? I am thinking of something similar to the get method of Python dictionaries!

Answer 1

You can be more pythonic replacing your try/except block with:

print(0 if len(query)==0 else query.iloc[0][0])

Explanation: len() applied to a pandas Dataframe returns the number of rows.

Update: as suggested in comments, query.empty this is more idiomatic and .iat is better for scalar lookups, hence:

print(0 if query.empty else query.iat[0,0])

Answer 2

There's no intrinsically better way than try / except . The rationale for iloc is indexing by integer positional location.

The behaviour and functionality is consistent with NumPy np.ndarray , Python list and other indexable objects. There's no direct way to index either the first value of a list or return 0 if the list is empty.

A slightly better way is to be explicit and catch IndexError only and use iat for accessing scalars by integer location. Moreover, you can index by row and column simultaneously :

try:
    print(query.iat[0, 0])
except IndexError:
    print(0)

Answer 3

You can probably use something like

next(iter(series, default))

For example, using your input

In [1]: 
import pandas as pd
mydict = {"col1":[1,2], "price":[1000,2000]}
df = pd.DataFrame(mydict)
df
Out[1]: 
   col1  price
0     1   1000
1     2   2000

and filtering on price > 2000, gives the default value (which we are setting to zero) since df.loc[mask] would be empty

In [2]: 
mask = (df['price']>2000)
next(iter(df.loc[mask]['col1']), 0)
Out[2]: 
0

The other cases work as expected. For example, filtering on price > 1500, gives 2

In [3]: 
mask = (df['price']>1500)
next(iter(df.loc[mask]['col1']), 0)
Out[3]: 
2

and filtering on price > 500 gives 1

In [4]: 
mask = (df['price']>500)
next(iter(df.loc[mask]['col1']), 0)
Out[4]: 
1