我在获取将函数应用于数据框的正确语法时遇到了问题。我正在尝试通过连接其他两列中的字符串并传入分隔符来在数据框中创建一个新列。我得到错误
TypeError: ("apply_join() missing 1 required positional argument: 'sep'", 'occurred at index cases')
如果我将 sep 添加到 apply_join() 函数调用，那也会失败:
File "unite.py", line 37, in unite
tibble_extra = df[cols].apply(apply_join, sep)
NameError: name 'sep' is not defined
import pandas as pd
from io import StringIO
tibble3_csv = """country,year,cases,population
Afghanistan,1999,745,19987071
Afghanistan,2000,2666,20595360
Brazil,1999,37737,172006362
Brazil,2000,80488,174504898
China,1999,212258,1272915272
China,2000,213766,1280428583"""
with StringIO(tibble3_csv) as fp:
tibble3 = pd.read_csv(fp)
print(tibble3)
def str_join_elements(x, sep=""):
assert type(sep) is str
return sep.join((str(xi) for xi in x))
def unite(df, cols, new_var, combine=str_join_elements):
def apply_join(x, sep):
joinstr = str_join(x, sep)
return pd.Series({new_var[i]:s for i, s in enumerate(joinstr)})
fixed_vars = df.columns.difference(cols)
tibble = df[fixed_vars].copy()
tibble_extra = df[cols].apply(apply_join)
return pd.concat([tibble, tibble_extra], axis=1)
table3_again = unite(tibble3, ['cases', 'population'], 'rate', combine=lambda x: str_join_elements(x, "/"))
print(table3_again)
最佳答案
当你有多个参数时使用lambda，即
df[cols].apply(lambda x: apply_join(x,sep),axis=1)
或者在args参数的帮助下传递参数即
df[cols].apply(apply_join,args=[sep],axis=1)
关于python - 将变量传递给 pandas 中的 apply()，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/46519139/