如何从 pandas DataFrame 的列中减去单个值

Question

i have one data frame suppose:

name age hb
ali  34  14
jex  16  13
aja  24  16
joy  23  12

i have a value say "5" that i want to substract from each member of column "hb"

new column could be:

hb
9
8
11
7

What is the best method to do this...

thanks and regards.

Answer 1

Simply subtract the scalar value from the pandas.Series , for numerical columns pandas would automatically broadcast the scalar value and subtract it from each element in the column. Example -

df['hb'] - 5 #Where `df` is your dataframe.

Demo -

In [43]: df
Out[43]:
  name  age  hb
0  ali   34  14
1  jex   16  13
2  aja   24  16
3  joy   23  12

In [44]: df['hb'] - 5
Out[44]:
0     9
1     8
2    11
3     7
Name: hb, dtype: int64

Answer 2

If you are using this:

df['hb'] - 5

you will get a new single column. But if you want to keep the rest then you have to use:

df['hb'] -= 5

Answer 3

您也可以使用 pandas.apply 函数来执行此操作

df.loc[:, "hb"] = df["hb"].apply(lambda x: x - 5)

Answer 4

If you want this subtraction to be saved in your DataFrame and avoid the old SettingWithCopyWarning , use loc :

df.loc["hb"] -= 5

Importantly, if you need to use multiple conditions for selecting a value range, put both into the loc call (chaining does not work for this):

df.loc[df.age==34,"hb"] -= 5

Answer 5

try this:

df["hb"] - 5

df["hb"] will select hb column and subtract 5 from it

Answer 6

eval lets you assign the new values directly to your existing column hb :

In [6]: df.eval("hb = hb - 5", inplace=True)

In [7]: df
Out[7]: 
  name  age  hb
0  ali   34   9
1  jex   16   8
2  aja   24  11
3  joy   23   7

Since inplace=True you don't need to assign it back to df .