Python pandas -> 按列名中的条件选择
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Python pandas -> select by condition in columns name
提问by CezarySzulc
I have df with column names: 'a', 'b', 'c' ... 'z'.
我有 df 列名：'a', 'b', 'c' ... 'z'。
print(my_df.columns)
Index(['a', 'b', 'c', ... 'y', 'z'],
dtype='object', name=0)
I have function which determine which columns should be displayed. For example:
我有确定应显示哪些列的功能。例如：
start = con_start()
stop = con_stop()
print(my_df.columns >= start) & (my_df <= stop)
My result is:
我的结果是：
[False False ... False False False False True True
True True False False]
My goal is display dataframe only with columns that satisfy my condition. If start = 'a' and stop = 'b', I want to have:
我的目标是仅使用满足我的条件的列显示数据框。如果开始 = 'a' 和停止 = 'b'，我想要：
0 a b
index1 index2
New York New York 0.000000 0.000000
California Los Angeles 207066.666667 214466.666667
Illinois Chicago 138400.000000 143633.333333
Pennsylvania Philadelphia 53000.000000 53633.333333
Arizona Phoenix 111833.333333 114366.666667
采纳答案by piRSquared
I want to make this robust and with as few assumptions as possible.
我想让它变得健壮，并尽可能少做假设。
option 1
use ilocwith array slicing
Assumptions:
选项 1与数组切片一起
使用假设：iloc
my_df.columns.is_uniqueevaluates toTrue- columns are already in order
my_df.columns.is_unique评估为True- 列已经排序
start = df.columns.get_loc(con_start())
stop = df.columns.get_loc(con_stop())
df.iloc[:, start:stop + 1]
option 2
use locwith boolean slicing
Assumptions:
选项 2与布尔切片一起
使用假设：loc
- column values are comparable
- 列值具有可比性
start = con_start()
stop = con_stop()
c = df.columns.values
m = (start <= c) & (stop >= c)
df.loc[:, m]
回答by Scott Boston
You can use slicing to achieve this with .loc:
您可以使用切片来通过 .loc 实现此目的：
df.loc[:,'a':'b']
回答by Petr Matuska
If your conditions are on a similar level of complexity as you shown in your example there is no need to use any additional function but just do filtering e.g.
如果您的条件与示例中所示的复杂程度相似，则无需使用任何其他功能，只需进行过滤即可
sweet_and_red_fruit = fruit[(fruit[sweet == 1) & (fruit["colour"] == "red")]
print(sweet_and_red_fruit)
OR if you want to just print
或者，如果您只想打印
print(fruit[(fruit[sweet == 1) & (fruit["colour"] == "red")])
回答by acidtobi
Generate a list of colums to display:
生成要显示的列列表：
cols = [x for x in my_df.columns if start <= x <= stop]
Use only these columns in your DataFrame:
在您的 DataFrame 中仅使用这些列：
my_df[cols]
回答by Binyamin Even
assuming resultis your [true/false]array and that lettersis [a...z]:
假设result是你的[true/false]数组，那letters就是[a...z]：
res=[letters[i] for i,r in enumerate(result) if r]
new_df=df[res]
