Question

我在循环中遇到列表推导问题。我想将一个列表中的项目添加到另一个列表中。

我在列表理解中使用ui和map class。

zip

输出：

from pandas import Series, DataFrame
import pandas
x = ['a', 'b', 'c', 'd', 'e']
y = [2, 3, 6, 7, 4]
ser = {'A': Series(x), 'B': Series(y)}
df = DataFrame(ser)
targets = df['A'].tolist()
df['A1999'] = [i + 1 for i in df['B']]
df['A2000'] = [i + 2 for i in df['B']]
df['A2001'] = [i + 3 for i in df['B']]
df['A2002'] = [i + 1.7 for i in df['B']]
df['A2003'] = [i + 1.1 for i in df['B']]
y = range(1999, 2004)
gaps = []
for t in targets:
    temp = []
    years = []
    for ele in y:
        target = 'A' + str(ele)
        new = df[target][df['A'] == t].tolist()
        temp.append(new)
        years.append(ele)
        gap = [list(map(list, zip(years, item))) for item in temp]
    gaps.append(gap)

我正在寻找的是：

[[[[1999, 3]], [[1999, 4]], [[1999, 5]], [[1999, 3.7000000000000002]],
  [[1999, 3.1000000000000001]]]...

如何修复列表理解以便添加[[[[1999, 3]], [[2000, 4]], [[2001, 5]], [[2002, 3.7000000000000002]], [[2003, 3.1000000000000001]]]...的所有年份，而不仅仅是第一个（即1999年）

我尝试了这个例子，但我认为我做的是同样的事情：

years list

或

gap = [[[years[i], x] for i, x in enumerate(y)] for y in temp]

Answer 1

用此

替换gap =

[list((x,y[0])) for x,y in zip(years,temp)]

使用列表理解从列表中映射项目？

1 个答案: