for column in df:
    print(df[column])

假设x因素，y标签(多列):

columns = [c for c in _df.columns if c in ['col1', 'col2','col3']]  #or '..c not in..'
_df.set_index(columns, inplace=True)
print( _df.index)

X, y =  _df.iloc[:,:4].values, _df.index.values

你可以使用iteritems():

for name, values in df.iteritems():
    print('{name}: {value}'.format(name=name, value=values[0]))

我有点晚了，但我是这么做的。的步骤:

创建所有列的列表 使用itertools获取x的组合 将每个结果R平方值与排除列列表一起附加到结果数据框架中 将结果DF按R平方的降序排序，看看哪个是最合适的。

这是我在DataFrame上使用的代码，称为aft_tmt。请随意推断您的用例。

import pandas as pd
# setting options to print without truncating output
pd.set_option('display.max_columns', None)
pd.set_option('display.max_colwidth', None)

import statsmodels.formula.api as smf
import itertools

# This section gets the column names of the DF and removes some columns which I don't want to use as predictors.
itercols = aft_tmt.columns.tolist()
itercols.remove("sc97")
itercols.remove("sc")
itercols.remove("grc")
itercols.remove("grc97")
print itercols
len(itercols)

# results DF
regression_res = pd.DataFrame(columns = ["Rsq", "predictors", "excluded"])

# excluded cols
exc = []

# change 9 to the number of columns you want to combine from N columns.
#Possibly run an outer loop from 0 to N/2?
for x in itertools.combinations(itercols, 9):
    lmstr = "+".join(x)
    m = smf.ols(formula = "sc ~ " + lmstr, data = aft_tmt)
    f = m.fit()
    exc = [item for item in x if item not in itercols]
    regression_res = regression_res.append(pd.DataFrame([[f.rsquared, lmstr, "+".join([y for y in itercols if y not in list(x)])]], columns = ["Rsq", "predictors", "excluded"]))

regression_res.sort_values(by="Rsq", ascending = False)

for column in df:
    print(df[column])

大多数答案都是通过列名，而不是直接迭代列。如果有多个具有相同名称的列，也会出现问题。如果你想迭代列，我建议:

for series in (df.iloc[:,i] for i in range(df.shape[1])):
   ...